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Linear Algebra: A second course, featuring proofs and Python

Sean Fitzpatrick

Appendix C Solutions to Selected Exercises

1 Vector spaces
1.2 Properties

Exercise 1.2.2.

1.2.2.a
Solution.
Suppose \(\uu+\vv=\uu+\ww\text{.}\) By adding \(-\uu\) on the left of each side, we obtain:
\begin{align*} -\uu+(\uu+\vv) \amp =-\uu+(\uu+\ww)\\ (-\uu+\uu)+\vv \amp =(-\uu+\uu)+\ww \quad \text{ by A3}\\ \zer+\vv \amp =\zer+\ww \quad \text{ by A5}\\ \vv \amp =\ww \quad \text{ by A4}\text{,} \end{align*}
which is what we needed to show.
1.2.2.b
Solution.
We have \(c\zer = c(\zer+\zer) = c\zer +c\zer\text{,}\) by A4 and S2, respectively. Adding \(-c\zer\) to both sides gives us
\begin{equation*} -c\zer+c\zer = -c\zer+(c\zer+c\zer)\text{.} \end{equation*}
Using associativity (A3), this becomes
\begin{equation*} -c\zer+c\zer = (-c\zer+c\zer)+c\zer\text{,} \end{equation*}
and since \(-c\zer+c\zer=\zer\) by A5, we get \(\zer =\zer+c\zer\text{.}\) Finally, we apply A4 on the right hand side to get \(\zer=c\zer\text{,}\) as required.
1.2.2.c
Solution.
Suppose there are two vectors \(\zer_1,\zer_2\) that act as additive identities. Then
\begin{align*} \zer_1 \amp = \zer_1+\zer_2 \quad \text{ since } \vv+\zer_2=\vv \text{ for any } \vv\\ \amp =\zer_2+\zer_1 \quad \text{ by axiom A2}\\ \amp \zer_2 \quad \text{ since } \vv+\zer_1=\vv \text{ for any } \vv \end{align*}
So any two vectors satisfying the property in A4 must, in fact, be the same.
1.2.2.d
Solution.
Let \(\vv\in V\text{,}\) and suppose there are vectors \(\ww_1,\ww_2\in V\) such that \(\vv+\ww_1=\zer\) and \(\vv+\ww_2=\zer\text{.}\) Then
\begin{align*} \ww_1 \amp = \ww_1+\zer \quad \text{ by A4}\\ \amp = \ww_1+(\vv+\ww_2) \quad \text{ by assumption}\\ \amp = (\ww_1+\vv)+\ww_2 \quad \text{ by A3}\\ \amp = (\vv+\ww_1)+\ww_2 \quad \text{ by A2}\\ \amp = \zer+\ww_2 \quad \text{ by assumption}\\ \amp \ww_2 \quad \text{ by A4}\text{.} \end{align*}

1.6 Linear Independence

Exercise 1.6.7.

Solution.
We set up a matrix and reduce:
Notice that this time we don’t get a unique solution, so we can conclude that these vectors are not independent. Furthermore, you can probably deduce from the above that we have \(2\vv_1+3\vv_2-\vv_3=\zer\text{.}\) Now suppose that \(\ww\in\spn\{\vv_1,\vv_2,\vv_3\}\text{.}\) In how many ways can we write \(\ww\) as a linear combination of these vectors?

Exercise 1.6.8.

Solution.
In each case, we set up the defining equation for independence, collect terms, and then analyze the resulting system of equations. (If you work with polynomials often enough, you can probably jump straight to the matrix. For now, let’s work out the details.)
Suppose
\begin{equation*} r(x^2+1)+s(x+1)+tx = 0\text{.} \end{equation*}
Then \(rx^2+(s+t)x+(r+s)=0=0x^2+0x+0\text{,}\) so
\begin{align*} r \amp =0\\ s+t \amp =0\\ r+s\amp =0\text{.} \end{align*}
And in this case, we don’t even need to ask the computer. The first equation gives \(r=0\) right away, and putting that into the third equation gives \(s=0\text{,}\) and the second equation then gives \(t=0\text{.}\)
Since \(r=s=t=0\) is the only solution, the set is independent.
Repeating for \(S_2\) leads to the equation
\begin{equation*} (r+2s+t)x^2+(-r+s+5t)x+(3r+5s+t)1=0. \end{equation*}
This gives us:

Exercise 1.6.9.

Solution.
We set a linear combination equal to the zero vector, and combine:
\begin{align*} a\bbm -1\amp 0\\0\amp -1\ebm +b\bbm 1\amp -1\\ -1\amp 1\ebm +c\bbm 1\amp 1\\1\amp 1\ebm +d \bbm 0\amp -1\\-1\amp 0\ebm = \bbm 0\amp 0\\ 0\amp 0\ebm\\ \bbm -a+b+c\amp -b+c-d\\-b+c-d\amp -a+b+c\ebm = \bbm 0\amp 0\\0\amp 0\ebm\text{.} \end{align*}
We could proceed, but we might instead notice right away that equations 1 and 4 are identical, and so are equations 2 and 3. With only two distinct equations and 4 unknowns, we’re certain to find nontrivial solutions.

1.7 Basis and dimension

Exercise 1.7.5.

Solution.
Let \(X=\bbm a\amp b\\c\amp d\ebm\text{.}\) Then \(AX = \bbm a+c\amp b+d\\0\amp 0\ebm\text{,}\) and \(XA = \bbm a\amp a\\c\amp c\ebm\text{,}\) so the condition \(AX=XA\) requires:
\begin{align*} a+c \amp = a\\ b+d \amp = a\\ 0 \amp = c\\ 0 \amp =c\text{.} \end{align*}
So \(c=0\text{,}\) in which case the first equation \(a=a\) is trivial, and we are left with the single equation \(a=b+d\text{.}\) Thus, our matrix \(X\) must be of the form
\begin{equation*} X = \bbm b+d \amp b\\0\amp d\ebm = b\bbm 1\amp 1\\0\amp 0\ebm + d\bbm 1\amp 0\\0\amp 1\ebm\text{.} \end{equation*}
Since the matrices \(\bbm 1\amp 1\\0\amp 0\ebm\) and \(\bbm 1\amp 0\\0\amp 1\ebm\) are not scalar multiples of each other, they must be independent, and therefore, they form a basis for \(U\text{.}\) (Why do we know these matrices span \(U\text{?}\))

Exercise 1.7.7.

1.7.7.a
Solution.
We need to show that the set is independent, and that it spans.
The set is independent if the equation
\begin{equation*} x(1,1,0)+y(1,0,1)+z(0,1,1)=(0,0,0) \end{equation*}
has \(x=y=z=0\) as its only solution. This equation is equivalent to the system
\begin{align*} x+y \amp =0\\ x+z \amp =0\\ y+z \amp =0\text{.} \end{align*}
We know that the solution to this system is unique if the coefficient matrix \(A = \bbm 1\amp 1\amp 0\\1\amp 0\amp 1\\0\amp 1\amp 1\ebm\) is invertible. Note that the columns of this matrix are vectors in our set.
We can determine invertibility either by showing that the RREF of \(A\) is the identity, or by showing that the determinant of \(A\) is nonzero. Either way, this is most easily done by the computer:
Our set of vectors is therefore linearly independent. Now, to show that it spans, we need to show that for any vector \((a,b,c)\text{,}\) the equation
\begin{equation*} x(1,1,0)+y(1,0,1)+z(0,1,1)=(a,b,c) \end{equation*}
has a solution. But we know that this system has the same coefficient matrix as the one above, and that existence of a solution again follows from invertibility of \(A\text{,}\) which we have already established.
Note that for three vectors in \(\R^3\text{,}\) once independence has been confirmed, span is automatic. We will soon see that this is not a coincidence.
1.7.7.b
Solution.
Based on what we learned from the first set, determining whether or not this set is a basis is equivalent to determining whether or not the matrix \(A\) whose columns consist of the vectors in the set is invertible. We form the matrix
\begin{equation*} A = \bbm -1\amp 1\amp 1\\1\amp -1\amp 1\\1\amp 1\amp -1\ebm \end{equation*}
and then check invertibility using the computer.
Since the determinant is nonzero, our set is a basis.

Exercise 1.7.10.

1.7.10.a
Solution.
By definition, \(U_1 = \spn \{1+x,x+x^2\}\text{,}\) and these vectors are independent, since neither is a scalar multiple of the other. Since there are two vectors in this basis, \(\dim U_1 = 2\text{.}\)
1.7.10.b
Solution.
If \(p(1)=0\text{,}\) then \(p(x)=(x-1)q(x)\) for some polynomial \(q\text{.}\) Since \(U_2\) is a subspace of \(P_2\text{,}\) the degree of \(q\) is at most 2. Therefore, \(q(x)=ax+b\) for some \(a,b\in\R\text{,}\) and
\begin{equation*} p(x) = (x-1)(ax+b) = a(x^2-x)+b(x-1)\text{.} \end{equation*}
Since \(p\) was arbitrary, this shows that \(U_2 = \spn\{x^2-x,x-1\}\text{.}\)
The set \(\{x^2-x,x-1\}\) is also independent, since neither vector is a scalar multiple of the other. Therefore, this set is a basis, and \(\dim U_2=2\text{.}\)
1.7.10.c
Solution.
If \(p(x)=p(-x)\text{,}\) then \(p(x)\) is an even polynomial, and therefore \(p(x)=a+bx^2\) for \(a,b\in\R\text{.}\) (If you didn’t know this it’s easily verified: if
\begin{equation*} a+bx+cx^2 = a+b(-x)+c(-x)^2\text{,} \end{equation*}
we can immediately cancel \(a\) from each side, and since \((-x)^2=x^2\text{,}\) we can cancel \(cx^2\) as well. This leaves \(bx=-bx\text{,}\) or \(2bx=0\text{,}\) which implies that \(b=0\text{.}\))
It follows that the set \(\{1,x^2\}\) spans \(U_3\text{,}\) and since this is a subset of the standard basis \(\{1,x,x^2\}\) of \(P_2\text{,}\) it must be independent, and is therefore a basis of \(U_3\text{,}\) letting us conclude that \(\dim U_3=2\text{.}\)

Exercise 1.7.14.

Solution.
By the previous theorem, we can form a basis by adding vectors from the standard basis
\begin{equation*} \left\{\bbm 1\amp 0\\0\amp 0\ebm, \bbm 0\amp 1\\0\amp 0\ebm, \bbm 0\amp 0\\1\amp 0\ebm, \bbm 0\amp 0\\0\amp 1\ebm\right\}\text{.} \end{equation*}
It’s easy to check that \(\bbm 1\amp 0\\0\amp 0\ebm\) is not in the span of \(\{\vv,\ww\}\text{.}\) To get a basis, we need one more vector. Observe that all three of our vectors so far have a zero in the \((2,1)\)-entry. Thus, \(\bbm 0\amp 0\\1\amp 0\ebm\) cannot be in the span of the first three vectors, and adding it gives us our basis.

Exercise 1.7.15.

Solution.
Again, we only need to add one vector from the standard basis \(\{1,x,x^2,x^3\}\text{,}\) and it’s not too hard to check that any of them will do.

1.8 New subspaces from old

Exercise 1.8.7.

1.8.7.a
Solution.
If \((x,y,z)\in U\text{,}\) then \(z=0\text{,}\) and if \((x,y,z)\in W\text{,}\) then \(x=0\text{.}\) Therefore, \((x,y,z)\in U\cap W\) if and only if \(x=z=0\text{,}\) so \(U\cap W = \{(0,y,0)\,|\, y\in\R\}\text{.}\)
1.8.7.b
Solution.
There are in fact infinitely many ways to do this. Three possible ways include:
\begin{align*} \vv \amp = (1,1,0)+(0,0,1) \\ \amp = (1,0,0)+(0,1,1)\\ \amp = \left(1,\frac12,0\right)+\left(0,\frac12,1\right)\text{.} \end{align*}

2 Linear Transformations
2.1 Definition and examples

Exercise 2.1.11.

Solution.
We need to find scalars \(a,b,c\) such that
\begin{equation*} 2-x+3x^2 = a(x+2)+b(1)+c(x^2+x)\text{.} \end{equation*}
We could set up a system and solve, but this time it’s easy enough to just work our way through. We must have \(c=3\text{,}\) to get the correct coefficient for \(x^2\text{.}\) This gives
\begin{equation*} 2-x+3x^2=a(x+2)+b(1)+3x^2+3x\text{.} \end{equation*}
Now, we have to have \(3x+ax=-x\text{,}\) so \(a=-4\text{.}\) Putting this in, we get
\begin{equation*} 2-x+3x^2=-4x-8+b+3x^2+3x\text{.} \end{equation*}
Simiplifying this leaves us with \(b=10\text{.}\) Finally, we find:
\begin{align*} T(2-x+3x^2) \amp = T(-4(x+2)+10(1)+3(x^2+x)) \\ \amp = -4T(x+2)+10T(1)+3T(x^2+x)\\ \amp = -4(1)+10(5)+3(0) = 46\text{.} \end{align*}

2.2 Kernel and Image

Exercise 2.2.10.

2.2.10.a
Solution.
We have \(T(0)=0\) since \(0^T=0\text{.}\) Using proerties of the transpose and matrix algebra, we have
\begin{equation*} T(A+B) = (A+B)-(A+B)^T = (A-A^T)+(B-B^T) = T(A)+T(B) \end{equation*}
and
\begin{equation*} T(kA) = (kA) - (kA)^T = kA-kA^T = k(A-A^T) = kT(A)\text{.} \end{equation*}
2.2.10.b
Solution.
It’s clear that if \(A^T=A\text{,}\) then \(T(A)=0\text{.}\) On the other hand, if \(T(A)=0\text{,}\) then \(A-A^T=0\text{,}\) so \(A=A^T\text{.}\) Thus, the kernel consists of all symmetric matrices.
2.2.10.c
Solution.
If \(B=T(A)=A-A^T\text{,}\) then
\begin{equation*} B^T = (A-A^T)^T = A^T-A = -B\text{,} \end{equation*}
so certainly every matrix in \(\im A\) is skew-symmetric. On the other hand, if \(B\) is skew-symmetric, then \(B=T(\frac12 B)\text{,}\) since
\begin{equation*} T\Bigl(\frac12 B\Bigr) = \frac12 T(B) = \frac12(B-B^T) = \frac12(B-(-B))= B\text{.} \end{equation*}

Exercise 2.2.17.

2.2.17.a
Solution.
Suppose \(T:V\to W\) is injective. Then \(\ker T = \{0\}\text{,}\) so
\begin{equation*} \dim V = 0 + \dim \im T \leq \dim W\text{,} \end{equation*}
since \(\im T\) is a subspace of \(W\text{.}\)
Conversely, suppose \(\dim V\leq \dim W\text{.}\) Choose a basis \(\{\vv_1,\ldots, \vv_m\}\) of \(V\text{,}\) and a basis \(\{\ww_1,\ldots, \ww_n\}\) of \(W\text{,}\) where \(m\leq n\text{.}\) By Theorem 2.1.8, there exists a linear transformation \(T:V\to W\) with \(T(\vv_i)=\ww_i\) for \(i=1,\ldots, m\text{.}\) (The main point here is that we run out of basis vectors for \(V\) before we run out of basis vectors for \(W\text{.}\)) This map is injective: if \(T(\vv)=\zer\text{,}\) write \(\vv=c_1\vv_1+\cdots + c_m\vv_m\text{.}\) Then
\begin{align*} \zer \amp = T(\vv)\\ \amp = T(c_1\vv_1+\cdots + c_m\vv_m)\\ \amp = c_1T(\vv_1)+\cdots + c_mT(\vv_m)\\ \amp = c_1\ww_1+\cdots +c_m\ww_m\text{.} \end{align*}
Since \(\{\ww_1,\ldots, \ww_m\}\) is a subset of a basis, it’s independent. Therefore, the scalars \(c_i\) must all be zero, and therefore \(\vv=\zer\text{.}\)
2.2.17.b
Solution.
Suppose \(T:V\to W\) is surjective. Then \(\dim \im T = \dim W\text{,}\) so
\begin{equation*} \dim V = \dim \ker T + \dim W \geq \dim W\text{.} \end{equation*}
Conversely, suppose \(\dim V\geq \dim W\text{.}\) Again, choose a basis \(\{\vv_1,\ldots, \vv_m\}\) of \(V\text{,}\) and a basis \(\{\ww_1,\ldots, \ww_n\}\) of \(W\text{,}\) where this time, \(m\geq n\text{.}\) We can define a linear transformation as follows:
\begin{equation*} T(\vv_1)=\ww_1,\ldots, T(\vv_n)=\ww_n, \text{ and } T(\vv_j) = \zer \text{ for } j>n. \end{equation*}
It’s easy to check that this map is a surjection: given \(\ww\in W\text{,}\) we can write it in terms of our basis as \(\ww=c_1\ww_1+\cdots + c_n\ww_n\text{.}\) Using these same scalars, we can define \(\vv=c_1\vv_1+\cdots + c_n\vv_n\in V\) such that \(T(\vv)=\ww\text{.}\)
Note that it’s not important how we define \(T(\vv_j)\) when \(j>n\text{.}\) The point is that this time, we run out of basis vectors for \(W\) before we run out of basis vectors for \(V\text{.}\) Once each vector in the basis of \(W\) is in the image of \(T\text{,}\) we’re guaranteed that \(T\) is surjective, and we can define the value of \(T\) on any remaining basis vectors however we want.

2.3 Isomorphisms, composition, and inverses
2.3.2 Composition and inverses

Exercise 2.3.7.

Solution.
Suppose we have linear maps \(U\xrightarrow{T} V\xrightarrow{S} W\text{,}\) and let \(\uu_1,\uu_2\in U\text{.}\) Then
\begin{align*} ST(\uu_1+\uu_2) \amp = S(T(\uu_1+\uu_2)) \\ \amp = S(T(\uu_1)+T(\uu_2))\\ \amp = S(T(\uu_1))+S(T(\uu_2)) \\ \amp = ST(\uu_1)+ST(\uu_2)\text{,} \end{align*}
and for any scalar \(c\text{,}\)
\begin{equation*} ST(c\uu_1) = S(T(c\uu_1))=S(cT(\uu_1)) = cS(T(\uu_1))=c(ST(\uu_1))\text{.} \end{equation*}

Exercise 2.3.9.

Solution.
Let \(\ww_1,\ww_2\in W\text{.}\) Then there exist \(\vv_1,\vv_2\in V\) with \(\ww_1=T(\vv_1), \ww_2=T(\vv_2)\text{.}\) We then have
\begin{align*} T^{-1}(\ww_1+\ww_2) \amp = T^{-1}(T(\vv_1)+T(\vv_2)) \\ \amp = T^{-1}(T(\vv_1+\vv_2))\\ \amp = \vv_1+\vv_2\\ \amp = T^{-1}(\ww_1)+T^{-1}(\ww_2)\text{.} \end{align*}
For any scalar \(c\text{,}\) we similarly have
\begin{equation*} T^{-1}(c\ww_1) = T^{-1}(cT(\vv_1))=T^{-1}(T(c\vv_1)) = c\vv_1 = cT^{-1}(\ww_1)\text{.} \end{equation*}

3 Orthogonality and Applications
3.1 Orthogonal sets of vectors
3.1.1 Basic definitions and properties

Exercise 3.1.5.

Solution.
This is simply an exercise in properties of the dot product. We have
\begin{align*} \len{\xx+\yy}^2 \amp = (\xx+\yy)\dotp (\xx+\yy) \\ \amp = \xx\dotp \xx+\xx\dotp\yy+\yy\dotp\xx+\yy\dotp\yy\\ \amp =\len{\xx}^2+2\xx\dotp\yy+\len{\yy}^2\text{.} \end{align*}

Exercise 3.1.6.

Solution.
If \(\xx=\zer\text{,}\) then the result follows immediately from the dot product formula in Definition 3.1.1. Conversely, suppose \(\xx\dotp \vv_i=0\) for each \(i\text{.}\) Since the \(\vv_i\) span \(\R^n\text{,}\) there must exist scalars \(c_1,c_2,\ldots, c_k\) such that \(\xx=c_1\vv_1+c_2\vv_2+\cdots+c_k\vv_k\text{.}\) But then
\begin{align*} \xx\dotp\xx \amp = \xx\dotp (c_1\vv_1+c_2\vv_2+\cdots+c_k\vv_k) \\ \amp = c_1(\xx\dotp \vv_1)+ c_2(\xx\dotp \vv_2)+\cdots +c_k(\xx\dotp \vv_k)\\ \amp = c_1(0)+c_2(0)+\cdots + c_k(0)=0\text{.} \end{align*}

3.1.2 Orthogonal sets of vectors

Exercise 3.1.13.

Solution.
Clearly, all three vectors are nonzero. To confirm the set is orthogonal, we simply compute dot products:
\begin{align*} (1,0,1,0)\dotp (-1,0,1,1)\amp =-1+0+1+0=0\\ (-1,0,1,1)\dotp (1,1,-1,2)\amp =-1+0-1+2=0\\ (1,0,1,0)\dotp (1,1,-1,2) \amp = 1+0-1+0=0\text{.} \end{align*}
To find a fourth vector, we proceed as follows. Let \(\xx=(a,b,c,d)\text{.}\) We want \(\xx\) to be orthogonal to the three vectors in our set. Computing dot products, we must have:
\begin{align*} (a,b,c,d)\dotp (1,0,1,0) \amp = a+c=0 \\ (a,b,c,d)\dotp (-1,0,1,1) \amp = -a+c+d=0 \\ (a,b,c,d)\dotp (1,1,-1,2) \amp = a+b-c+2d=0\text{.} \end{align*}
This is simply a homogeneous system of three equations in four variables. Using the Sage cell below, we find that our vector must satisfy \(a=\frac12 d, b = -3d, c=-\frac12 d\text{.}\)
One possible nonzero solution is to take \(d=2\text{,}\) giving \(\xx=(1,-6,-1,2)\text{.}\) We’ll leave the verification that this vector works as an exercise.

Exercise 3.1.18.

Solution.
We compute
\begin{align*} \left(\frac{\vv\dotp\xx_1}{\len{\xx_1}^2}\right)\xx_1 \amp +\left(\frac{\vv\dotp\xx_2}{\len{\xx_2}^2}\right)\xx_2 +\left(\frac{\vv\dotp\xx_3}{\len{\xx_3}^2}\right)\xx_3\\ \amp = \frac{4}{2}\xx_1+\frac{-9}{3}\xx_2+\frac{-28}{7}\xx_3\\ \amp = 2(1,0,1,0)-3(-1,0,1,1)-4(1,1,-1,2)\\ \amp = (1,-4,3,-11) = \vv\text{,} \end{align*}
so \(\vv\in\spn\{\xx_1,\xx_2,\xx_3\}\text{.}\)
On the other hand, repeating the same calculation with \(\ww\text{,}\) we find
\begin{align*} \left(\frac{\vv\dotp\xx_1}{\len{\xx_1}^2}\right)\xx_1 \amp +\left(\frac{\vv\dotp\xx_2}{\len{\xx_2}^2}\right)\xx_2 +\left(\frac{\vv\dotp\xx_3}{\len{\xx_3}^2}\right)\xx_3\\ \amp =\frac12 (1,0,1,0)-\frac53 (-1,0,1,1) +\frac47 (1,1,-1,2)\\ \amp = \left(\frac{73}{42},\frac47,-\frac{115}{42},-\frac{11}{21}\right)\neq \ww\text{,} \end{align*}
so \(\ww\notin\spn\{\xx_1,\xx_2,\xx_3\}\text{.}\)
Soon, we’ll see that the quantity we computed when showing that \(\ww\notin\spn\{\xx_1,\xx_2,\xx_3\}\) is, in fact, the orthogonal projection of \(\ww\) onto the subspace \(\spn\{\xx_1,\xx_2,\xx_3\}\text{.}\)

4 Diagonalization
4.2 Diagonalization of symmetric matrices

Exercise 4.2.1.

Solution.
Take \(\xx=\mathbf{e}_i\) and \(\yy=\mathbf{e}_j\text{,}\) where \(\{\mathbf{e}_1,\ldots, \mathbf{e}_n\}\) is the standard basis for \(\R^n\text{.}\) Then with \(A = [a_{ij}]\) we have
\begin{equation*} a_{ij} =\mathbf{e}_i\dotp(A\mathbf{e}_j) = (A\mathbf{e}_i)\dotp \mathbf{e}_j = a_{ji}\text{,} \end{equation*}
which shows that \(A^T=A\text{.}\)

4.4 Diagonalization of complex matrices
4.4.2 Complex matrices

Exercise 4.4.8.

Solution.
We have \(\bar{A}=\bbm 4\amp 1+i\amp -2-3i\\1-i\amp 5 \amp -7i\\-2+3i\amp 7i\amp -4\ebm\text{,}\) so
\begin{equation*} A^H = (\bar{A})^T = \bbm 4\amp 1-i\amp -2+3i\\1+i\amp 5\amp 7i\\-2-3i\amp -7i\amp -4\ebm = A\text{,} \end{equation*}
and
\begin{align*} BB^H \amp =\frac14\bbm 1+i\amp \sqrt{2}\\1-i\amp\sqrt{2}i\ebm\bbm 1-i\amp 1+i\\\sqrt{2}\amp-\sqrt{2}i\ebm \\ \amp =\frac14\bbm (1+i)(1-i)+2\amp (1+i)(1+i)-2i\\(1-i)(1-i)+2i\amp (1-i)(1+i)+2\ebm\\ \amp =\frac14\bbm 4\amp 0\\0\amp 4\ebm = \bbm 1\amp 0\\0\amp 1\ebm\text{,} \end{align*}
so that \(B^H = B^{-1}\text{.}\)

Exercise 4.4.12.

Solution.
Confirming that \(A^H=A\) is almost immediate. We will use the computer below to compute the eigenvalues and eigenvectors of \(A\text{,}\) but it’s useful to attempt this at least once by hand. We have
\begin{align*} \det(zI-A) \amp = \det\bbm z-4 \amp -3+i\\-3-i\amp z-1\ebm\\ \amp (z-4)(z-1)-(-3-i)(-3+i)\\ \amp z^2-5z+4-10\\ \amp (z+1)(z-6)\text{,} \end{align*}
so the eigenvalues are \(\lambda_1=-1\) and \(\lambda_2=6\text{,}\) which are both real, as expected.
Finding eigenvectors can seem trickier than with real numbers, mostly because it is no longer immediately apparent when one row or a matrix is a multiple of another. But we know that the rows of \(A-\lambda I\) must be parallel for a \(2\times 2\) matrix, which lets proceed nonetheless.
For \(\lambda_1=-1\text{,}\) we have
\begin{equation*} A + I =\bbm 5 \amp 3-i\\3+i\amp 2\ebm\text{.} \end{equation*}
There are two ways one can proceed from here. We could use row operations to get to the reduced row-echelon form of \(A\text{.}\) If we take this approach, we multiply row 1 by \(\frac15\text{,}\) and then take \(-3-i\) times the new row 1 and add it to row 2, to create a zero, and so on.
Easier is to realize that if we haven’t made a mistake calculating our eigenvalues, then the above matrix can’t be invertible, so there must be some nonzero vector in the kernel. If \((A+I)\bbm a\\b\ebm=\bbm0\\0\ebm\text{,}\) then we must have
\begin{equation*} 5a+(3-i)b=0\text{,} \end{equation*}
when we multiply by the first row of \(A\text{.}\) This suggests that we take \(a=3-i\) and \(b=-5\text{,}\) to get \(\zz = \bbm 3-i\\-5\ebm\) as our first eigenvector. To make sure we’ve done things correctly, we multiply by the second row of \(A+I\text{:}\)
\begin{equation*} (3+i)(3-i)+2(-5) = 10-10 = 0\text{.} \end{equation*}
Success! Now we move onto the second eigenvalue.
For \(\lambda_2=6\text{,}\) we get
\begin{equation*} A-6I = \bbm -2\amp 3-i\\3+i\amp -5\ebm\text{.} \end{equation*}
If we attempt to read off the answer like last time, the first row of \(A-6I\) suggests the vector \(\ww = \bbm 3-i\\2\ebm\text{.}\) Checking the second row to confirm, we find:
\begin{equation*} (3+i)(3-i)-5(2) = 10-10=0\text{,} \end{equation*}
as before.
Finally, we note that
\begin{equation*} \langle \zz, \ww\rangle = (3-i)\overline{(3-i)}+(-5)(2) = (3-i)(3+i)-10 = 0\text{,} \end{equation*}
so the two eigenvectors are orthogonal, as expected. We have
\begin{equation*} \len{\zz}=\sqrt{10+25}=\sqrt{35} \quad \text{ and } \quad \len{\ww}=\sqrt{10+4}=\sqrt{14}\text{,} \end{equation*}
so our orthogonal matrix is
\begin{equation*} U = \bbm \frac{3-i}{\sqrt{35}}\amp \frac{3-i}{\sqrt{14}}\\-\frac{5}{\sqrt{35}}\amp \frac{2}{\sqrt{14}}\ebm\text{.} \end{equation*}
With a bit of effort, we can finally confirm that
\begin{equation*} U^HAU = \bbm -1\amp 0\\0\amp 6\ebm\text{,} \end{equation*}
as expected.

5 Change of Basis
5.1 The matrix of a linear transformation

Exercise 5.1.2.

Solution.
It’s clear that \(C_B(\zer)=\zer\text{,}\) since the only way to write the zero vector in \(V\) in terms of \(B\) (or, indeed, any independent set) is to set all the scalars equal to zero.
If we have two vectors \(\vv,\ww\) given by
\begin{align*} \vv \amp = a_1\mathbf{e}_1+\cdots + a_n\mathbf{e}_n \\ \ww \amp = b_1\mathbf{e}_1+\cdots + b_n\mathbf{e}_n\text{,} \end{align*}
then
\begin{equation*} \vv+\ww = (a_1+b_1)\mathbf{e}_1+\cdots + (a_n+b_n)\mathbf{e}_n\text{,} \end{equation*}
so
\begin{align*} C_B(\vv+\ww) \amp = \bbm a_1+b_1\\\vdots \\ a_n+b_n\ebm \\ \amp = \bbm a_1\\\vdots\\a_n\ebm +\bbm b_1\\\vdots \\b_n\ebm\\ \amp = C_B(\vv)+C_B(\ww)\text{.} \end{align*}
Finally, for any scalar \(c\text{,}\) we have
\begin{align*} C_B(c\vv) \amp = C_B((ca_1)\mathbf{e}_1+\cdots +(ca_n)\mathbf{e}_n)\\ \amp = \bbm ca_1\\\vdots \\ca_n\ebm\\ \amp =c\bbm a_1\\\vdots \\a_n\ebm\\ \amp =cC_B(\vv)\text{.} \end{align*}
This shows that \(C_B\) is linear. To see that \(C_B\) is an isomorphism, we can simply note that \(C_B\) takes the basis \(B\) to the standard basis of \(\R^n\text{.}\) Alternatively, we can give the inverse: \(C_B^{-1}:\R^n\to V\) is given by
\begin{equation*} C_B^{-1}\bbm c_1\\\vdots \\c_n\ebm = c_1\mathbf{e}_1+\cdots +c_n\mathbf{e}_n\text{.} \end{equation*}

Exercise 5.1.4.

Solution.
We have
\begin{align*} T(1) \amp = (1,0) = 1(1,0)+0(1,-1) \\ T(1-x) \amp= (1,-2) = -1(1,0)+2(1,-1) \\ T((1-x)^2) = T(1-2+x^2) \amp = (2, -4) = -2(1,0)+4(1,-1)\text{.} \end{align*}
Thus,
\begin{align*} M_{DB}(T) \amp = \bbm C_D(T(1))\amp C_D(T(1-x)) \amp C_D(T((1-x)^2))\ebm\\ \amp = \bbm 1\amp -1\amp -2\\0\amp 2\amp 4\ebm\text{.} \end{align*}
To confirm, note that
\begin{align*} M_{DB}(T)C_B\amp (a+bx+cx^2)\\ \amp = M_{DB}(T)C_B((a+b+c)-(b+2c)(1-x)+c(1-x)^2)\\ \amp = \bbm 1\amp -1\amp -2\\0\amp 2\amp 4\ebm\bbm a+b+c\\ -b-2c\\c\ebm\\ \amp \bbm (a+b+c)+(b+2c)-2c\\0-2(b+2c)+4c\ebm = \bbm a+2b+c\\-2b\ebm\text{,} \end{align*}
while on the other hand,
\begin{align*} C_D(T(a+bx+cx^2)) \amp= C_D(a+c,2b)\\ \amp = C_D((a+2b+c)(1,0)-2b(1,-1))\\ \amp = \bbm a+2b+c\\-2b\ebm\text{.} \end{align*}

Exercise 5.1.8.

Solution.
We must first write our general input in terms of the given basis. With respect to the standard basis
\begin{equation*} B_0 = \left\{\bbm 1\amp 0\\0\amp 0\ebm, \bbm 0\amp 1\\0\amp 0\ebm, \bbm 0\amp 0\\1\amp 0\ebm, \bbm 0\amp 0\\0\amp 1\ebm\right\}\text{,} \end{equation*}
we have the matrix \(P = \bbm 1\amp 0\amp 0\amp 1\\0\amp 1\amp 1\amp 0\\0\amp 0\amp 1\amp 0\\0\amp 1\amp 0\amp 1\ebm\text{,}\) representing the change from the basis \(B\) the basis \(B_0\text{.}\) The basis \(D\) of \(P_2(\R)\) is already the standard basis, so we need the matrix \(M_{DB}(T)P^{-1}\text{:}\)
For a matrix \(X = \bbm a\amp b\\c\amp d\ebm\) we find
\begin{equation*} M_{DB}(T)P^{-1}C_{B_0}(X)=\bbm 2\amp -2\amp 2\amp 1\\0\amp 3\amp -8\amp 1\\-1\amp 1\amp 2\amp -1\ebm\bbm a\\b\\c\\d\ebm = \bbm 2a-2b+2c+d\\3b-8c+d\\-a+b+2c-d\ebm\text{.} \end{equation*}
But this is equal to \(C_D(T(X))\text{,}\) so
\begin{align*} T\left(\bbm a\amp b\\c\amp d\ebm\right) \amp = C_D^{-1}\bbm 2a-2b+2c+d\\3b-8c+d\\-a+b+2c-d\ebm\\ \amp = (2a-2b+2c+d)+(3b-8c+d)x+(-a+b+2c-d)x^2\text{.} \end{align*}

5.2 The matrix of a linear operator

Exercise 5.2.4.

Solution.
With respect to the standard basis, we have
\begin{equation*} M_0=M_{B_0}(T) = \bbm 3\amp -2\amp 4\\1\amp -5\amp 0\\0\amp 2\amp -7\ebm\text{,} \end{equation*}
and the matrix \(P\) is given by \(P = \bbm 1\amp 3\amp 1\\2\amp -1\amp 2\\0\amp 2\amp-5\ebm\text{.}\) Thus, we find
\begin{equation*} M_B(T)=P^{-1}M_0P=\bbm 9\amp 56\amp 36\\7\amp 15\amp 15\\-10\amp -46\amp -33\ebm\text{.} \end{equation*}

5.7 Jordan Canonical Form

Exercise 5.7.7.

Solution.
With respect to the standard basis of \(\R^4\text{,}\) the matrix of \(T\) is
\begin{equation*} M = \bbm 1\amp 1\amp 0\amp 0\\0\amp 1\amp 0\amp 0\\0\amp -1\amp 2\amp 0\\1\amp -1\amp 1\amp 1\ebm\text{.} \end{equation*}
We find (perhaps using the Sage cell provided below, and the code from the example above) that
\begin{equation*} c_T(x)=(x-1)^3(x-2)\text{,} \end{equation*}
so \(T\) has eigenvalues \(1\) (of multiplicity \(3\)), and \(2\) (of multiplicity \(1\)).
We tackle the repeated eigenvalue first. The reduced row-echelon form of \(M-I\) is given by
\begin{equation*} R_1 = \bbm 1\amp 0\amp 0\amp 0\\0\amp 1\amp 0\amp 0\\0\amp 0\amp 1\amp 0\\0\amp 0\amp 0\amp 0\ebm\text{,} \end{equation*}
so
\begin{equation*} E_1(M) = \spn\{\xx_1\}, \text{ where } \xx_1 = \bbm 0\\0\\0\\1\ebm\text{.} \end{equation*}
We now attempt to solve \((M-I)\xx=\xx_1\text{.}\) We find
\begin{equation*} \left(\begin{matrix}0\amp 1\amp 0\amp 0\\0\amp 0\amp 0\amp 0\\0\amp -1\amp 1\amp 0\\1\amp -1\amp 1\amp 0\end{matrix}\right|\left.\begin{matrix}0\\0\\0\\1\end{matrix}\right) \xrightarrow{\text{RREF}} \left(\begin{matrix} 1\amp 0\amp 0\amp 0\\0\amp 1\amp 0\amp 0\\0\amp 0\amp 1\amp 0\\0\amp 0\amp 0\amp 0\end{matrix}\right|\left.\begin{matrix}1\\0\\0\\0\end{matrix}\right)\text{,} \end{equation*}
so \(\xx = t\xx_1+\xx_2\text{,}\) where \(\xx_2 = \bbm 1\\0\\0\\0\ebm\text{.}\) We take \(\xx_2\) as our first generalized eigenvector. Note that \((M-I)^2\xx_2 = (M-I)\xx_1=\zer\text{,}\) so \(\xx_2\in \nll (M-I)^2\text{,}\) as expected.
Finally, we look for an element of \(\nll (M-I)^3\) of the form \(\xx_3\text{,}\) where \((M-I)\xx_3=\xx_2\text{.}\) We set up and solve the system \((M-I)\xx=\xx_2\) as follows:
\begin{equation*} \left(\begin{matrix}0\amp 1\amp 0\amp 0\\0\amp 0\amp 0\amp 0\\0\amp -1\amp 1\amp 0\\1\amp -1\amp 1\amp 0\end{matrix}\right|\left.\begin{matrix}1\\0\\0\\0\end{matrix}\right) \xrightarrow{\text{RREF}} \left(\begin{matrix} 1\amp 0\amp 0\amp 0\\0\amp 1\amp 0\amp 0\\0\amp 0\amp 1\amp 0\\0\amp 0\amp 0\amp 0\end{matrix}\right|\left.\begin{matrix}0\\1\\1\\0\end{matrix}\right)\text{,} \end{equation*}
so \(\xx = t\xx_1+\xx_3\text{,}\) where \(\xx_3 =\bbm 0\\1\\1\\0\ebm\text{.}\)
Finally, we deal with the eigenvalue \(2\text{.}\) The reduced row-echelon form of \(M-2I\) is
\begin{equation*} R_2 = \bbm 1\amp 0\amp 0\amp 0\\0\amp 1\amp 0\amp 0\\0\amp 0\amp 1\amp -1\\0\amp 0\amp 0\amp 0\ebm\text{,} \end{equation*}
so
\begin{equation*} E_2(M) = \spn\{\yy\}, \text{ where } \yy = \bbm 0\\0\\1\\1\ebm\text{.} \end{equation*}
Our basis of column vectors is therefore \(B=\{\xx_1,\xx_2,\xx_3,\yy\}\text{.}\) Note that by design,
\begin{align*} M\xx_1 \amp =\xx_1\\ M\xx_2 \amp =\xx_1+\xx_2\\ M\xx_3 \amp= \xx_2+\xx_3\\ M\yy \amp = 2\yy\text{.} \end{align*}
The corresponding Jordan basis for \(\R^4\) is
\begin{equation*} \{(0,0,0,1),(1,0,0,0),(0,1,1,0),(0,0,1,1)\}\text{,} \end{equation*}
and with respect to this basis, we have
\begin{equation*} M_B(T) = \bbm 1\amp 1\amp 0\amp 0\\ 0\amp 1\amp 1\amp 0\\ 0\amp 0\amp 1\amp 0\\ 0\amp 0\amp 0\amp 2\ebm\text{.} \end{equation*}
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