Lecture 2 Practice Set

$[y]_{\mathcal B}=(3,3)^{\mathsf T}$

Check it forwards: $3(1,1)+3(1,-1)=(6,0)$. Or read it off: $B^{-1}(6,0)=\tfrac12(6,6)=(3,3)$.

Here $B=\begin{bmatrix}1&1\\1&-1\end{bmatrix}$, the matrix whose columns are the vectors of $\mathcal B$ (not the $B$ of problems 2, 3 and 5), and $B^{-1}=\tfrac12B$.

Column $j$ is $b_j$ written in $C$. For $b_1=(1,3)^{\mathsf T}=\alpha(1,1)^{\mathsf T}+\beta(-1,1)^{\mathsf T}$, solve $\alpha-\beta=1$ and $\alpha+\beta=3$: $\alpha=2$, $\beta=1$. For $b_2=(0,2)^{\mathsf T}$: $\alpha=\beta=1$.

$$P_{C\leftarrow B}=\begin{bmatrix}2&1\\1&1\end{bmatrix}$$

Line the labels up: $P_{C\leftarrow B}P_{D\leftarrow C}$ has $B$ meeting $D$ on the inside, and they do not match, so the product converts nothing. It is a perfectly good matrix, $\begin{bmatrix}4&1\\3&1\end{bmatrix}$, and it is not the one you want. The right order lets the two $C$'s meet: $P_{D\leftarrow B}=P_{D\leftarrow C}P_{C\leftarrow B}=\begin{bmatrix}2&1\\5&3\end{bmatrix}$. Square shapes will never warn you here; the inner labels are the check.

$[T]_{\mathcal E}=\begin{pmatrix}2&1\\1&2\end{pmatrix}$

$$\begin{gathered}[x]_B=B^{-1}[x]_E\\[2pt][Tx]_B=[T]_B[x]_B\\[2pt][Tx]_E=B[Tx]_B\\[3pt]\boxed{[T]_E=B[T]_BB^{-1}}\end{gathered}$$

The three lines: into $B$, act, back out; together they are the box.

With the same $B$ as in problem 1: $B[T]_{\mathcal B}B^{-1}=\begin{bmatrix}3&1\\3&-1\end{bmatrix}\cdot\tfrac12\begin{bmatrix}1&1\\1&-1\end{bmatrix}=\begin{bmatrix}2&1\\1&2\end{bmatrix}$.

A $C$-column needs $P_{B\leftarrow C}=P_{C\leftarrow B}^{-1}$. Here $\det P_{C\leftarrow B}=2\cdot1-1\cdot1=1\neq0$, so the inverse exists; swap the diagonal entries and negate the other two:

$$[x]_B=\begin{bmatrix}1&-1\\-1&2\end{bmatrix}\begin{bmatrix}3\\1\end{bmatrix} =\begin{bmatrix}2\\-1\end{bmatrix}$$

$t=\arccos\tfrac5{13}\approx67.38^\circ$. The first column says where the first axis lands, so $\cos t=\tfrac5{13}$ and $\sin t=\tfrac{12}{13}$. Both are positive, so $t$ lies between $0^\circ$ and $90^\circ$. The sign of $\sin t$ is the step not to skip: $\arccos$ alone cannot tell a turn through $t$ from a turn through $-t$.

(a) A rotation. Each column against itself is $\tfrac{16}{25}+\tfrac{9}{25}=1$; the two columns against each other are $\tfrac{12}{25}-\tfrac{12}{25}=0$. So $Q_1^{\mathsf T}Q_1=I$, and $\det Q_1=\tfrac{16+9}{25}=1$.

(b) Not a rotation, though it passes $Q_2^{\mathsf T}Q_2=I$: its columns are unit and perpendicular. It fails $\det Q_2=1$, since $\det Q_2=\tfrac{3(-3)-4(4)}{25}=-1$.

(c) Not a rotation, and it fails the other condition: $\det Q_3=1$, but $Q_3^{\mathsf T}Q_3=\begin{bmatrix}4&0\\0&\tfrac14\end{bmatrix}\neq I$.

$[x]_R=R^{\mathsf T}x=\tfrac15\bigl(3(0)+4(5),\;-4(0)+3(5)\bigr)^{\mathsf T}=(4,3)^{\mathsf T}$. The same point of the plane, with a new address.

This is the coordinate map $B^{-1}x$ of the first section, with the basis matrix now $R$; its inverse is free because $R^{\mathsf T}R=I$. The usual wrong turn is $Rx=(-4,3)^{\mathsf T}$: a different arrow, also of length $5$, read against the same axes as before.

$R_{ac}(2,1)^{\mathsf T}=(-1,2)^{\mathsf T}$, with

$$R_{ac}=R_{ab}R_{bc}=\tfrac1{25}\begin{bmatrix}12-12&-9-16\\16+9&-12+12\end{bmatrix} =\begin{bmatrix}0&-1\\1&0\end{bmatrix}.$$

The labels do the bookkeeping: in $R_{ab}R_{bc}$ the inner $b$'s meet and vanish, leaving $a$ from $c$, and the right-hand factor acts first — on the $c$-column, which is what you were handed.

No. Orthogonal means square with orthonormal columns, and $Q$ is $3\times2$; only a square matrix can have an inverse at all. $Q^{\mathsf T}$ undoes $Q$ from the left, not from the right:

$$QQ^{\mathsf T}=\begin{bmatrix}9/25&12/25&0\\12/25&16/25&0\\0&0&1\end{bmatrix}\neq I_3.$$

$\det M=(3)(-1)-(1)(2)=-3-2=-5$.

One factor works for every region, so the image has area $|\det M|\cdot 2=5\cdot2=10$. area 10

The likely wrong turn is to report the area as $-10$: an area is never negative, and the minus sign in $\det M$ says only that $M$ turns the plane over.

(a) Upper triangular, so the determinant is the product of the diagonal: $2\cdot3\cdot(-1)=-6$.

(b) Lower triangular. Its transpose is upper triangular with the same diagonal, and $\det L^{\mathsf T}=\det L$, so the same rule applies: $\det L=4\cdot(-2)\cdot0=0$.

$\det K=4$, so no.

Expanding along the first row:

$2(4-1)-1(2-0)+0=6-2=4$, so $Kx=b$ has exactly one solution for every $b$.

Or eliminate — subtracting a multiple of one row leaves $\det$ alone: $r_2-\tfrac12r_1$, then $r_3-\tfrac23r_2$, leaves pivots $2,\tfrac32,\tfrac43$, product $4$.

(a) $\det(AB)=\det A\cdot\det B=-6$.

(b) $\det(A^{-1})=\tfrac13$. From $AA^{-1}=I$: $\det A\cdot\det(A^{-1})=\det I=1$.

(c) $\det(A^{\mathsf T}B^{2})=\det(A^{\mathsf T})\cdot(\det B)^{2}=3\cdot4=12$. Transposing changes nothing, and the square contributes its factor twice.

(d) Cannot be determined. Two pairs with the stated determinants:

• $A=\operatorname{diag}(3,1,1)$, $B=\operatorname{diag}(-2,1,1)$: $A+B=\operatorname{diag}(1,2,2)$, determinant $4$.
• $A=\operatorname{diag}(1,3,1)$, $B=\operatorname{diag}(-2,1,1)$: $A+B=\operatorname{diag}(-1,4,2)$, determinant $-8$.

Exactly the even $n$. The rule $\det(cA)=c^{n}\det A$ with $c=-1$ gives $\det(-A)=(-1)^{n}\det A$. For odd $n$ the two agree only in the one case $\det A=0$, which is not every $A$.

The likely wrong turn is $\det(-A)=-\det A$, as if the determinant were linear in the matrix; but $-A$ scales all $n$ directions of the box by $-1$, so that factor comes out $n$ times.

Testing is one matrix–vector product each.

• $A(2,3)^{\mathsf T}=(6,9)^{\mathsf T}=3(2,3)^{\mathsf T}$. yes, $\lambda_1=3$
• $A(1,-1)^{\mathsf T}=(-2,2)^{\mathsf T}=-2(1,-1)^{\mathsf T}$. yes, $\lambda_2=-2$
• $A(1,0)^{\mathsf T}=(0,3)^{\mathsf T}$, which is not any multiple of $(1,0)^{\mathsf T}$. no

A negative eigenvalue is allowed: on $\operatorname{span}\{(1,-1)^{\mathsf T}\}$ the map is reversal through the origin together with a doubling.

Write $A$ for this matrix; it is not the $A$ of problem 1.

3 and 5: $\det(A-\lambda I)=(3-\lambda)(5-\lambda)$, and a triangular determinant is the product of the diagonal.

For $\lambda=3$, $v=(1,0)$. For $\lambda=5$ it is $(7,2)$, not $(0,1)$: $(A-5I)(7,2)=(-14+14,\,0)=0$.

Each eigenspace is the null space $\operatorname{Null}(B-\lambda_i I)$. Since $\lambda_i$ is an eigenvalue, that matrix is singular, so one row must vanish.

$\lambda_1=5$: $B-5I=\begin{bmatrix}-1&1\\2&-2\end{bmatrix}\to\begin{bmatrix}-1&1\\0&0\end{bmatrix}$ by $r_2\to r_2+2r_1$: one pivot, one free. Read $-x_1+x_2=0$, so $v_1=(1,1)^{\mathsf T}$; check $Bv_1=(5,5)^{\mathsf T}=5v_1$.

$\lambda_2=2$: $B-2I=\begin{bmatrix}2&1\\2&1\end{bmatrix}\to\begin{bmatrix}2&1\\0&0\end{bmatrix}$ by $r_2\to r_2-r_1$: one pivot, one free. Read $2x_1+x_2=0$, so $v_2=(1,-2)^{\mathsf T}$; check $Bv_2=(2,-4)^{\mathsf T}=2v_2$.

$\lambda_1=0$ and $\lambda_2=5$. The eigenvalues multiply to $\det C=0$, so one of them is $0$; they add to $\operatorname{tr}C=5$, so the other is $5$. The characteristic polynomial agrees: $\lambda^{2}-(\operatorname{tr}C)\lambda+\det C=\lambda^{2}-5\lambda=\lambda(\lambda-5)$.

$Q-\lambda I=\begin{bmatrix}-\lambda&-1\\1&-\lambda\end{bmatrix}$, so the main-diagonal product minus the anti-diagonal product is

$$\det(Q-\lambda I)=\lambda^{2}-(-1)(1)=\lambda^{2}+1=0$$

$\Delta=0^{2}-4(1)(1)=-4<0$: no real root, so no real eigenvector. Over $\mathbb C$ the roots are $\lambda_{1,2}=\pm i$.

A different matrix. The swapped $V'=\begin{bmatrix}1&2\\1&1\end{bmatrix}$ has $\det V'=-1$, so

$$V'\Lambda=\begin{bmatrix}2&-2\\2&-1\end{bmatrix},\qquad V'\Lambda V'^{-1}=\begin{bmatrix}2&-2\\2&-1\end{bmatrix}\begin{bmatrix}-1&2\\1&-1\end{bmatrix} =\begin{bmatrix}-4&6\\-3&5\end{bmatrix}$$

The right order gives $M=\begin{bmatrix}5&-6\\3&-4\end{bmatrix}$. Same two eigenvalues, but now $2$ belongs to $(1,1)^{\mathsf T}$ and $-1$ to $(2,1)^{\mathsf T}$. Reordering is allowed only if the columns of $V$ and the diagonal of $\Lambda$ move together.

$A^{k}=V\Lambda^{k}V^{-1}$, and $\Lambda^{k}$ raises each diagonal entry to the power $k$. With $V=\begin{bmatrix}1&0\\1&1\end{bmatrix}$ and $\det V=1$, $V^{-1}=\begin{bmatrix}1&0\\-1&1\end{bmatrix}$:

$$V\Lambda^{k}=\begin{bmatrix}3^{k}&0\\3^{k}&2^{k}\end{bmatrix},\qquad A^{k}=(V\Lambda^{k})V^{-1}=\begin{bmatrix}3^{k}&0\\3^{k}-2^{k}&2^{k}\end{bmatrix}$$

At $k=2$ it gives $\begin{bmatrix}9&0\\5&4\end{bmatrix}$, and $A\cdot A$ gives the same: the $(2,1)$ entry is $1\cdot3+2\cdot1=5$.

It approaches $c_1v_1=2(1,1)^{\mathsf T}$. Split $x_0$ into modes: $c_1(1,1)+c_2(2,-1)=(4,1)$, and subtracting the two equations gives $3c_2=3$, so $c_2=1$ and $c_1=2$:

$$x_k=2\cdot2^{k}\begin{bmatrix}1\\1\end{bmatrix}+(-1)^{k}\begin{bmatrix}2\\-1\end{bmatrix},\qquad \frac{x_k}{2^{k}}=2\begin{bmatrix}1\\1\end{bmatrix}+\left(-\tfrac12\right)^{k}\begin{bmatrix}2\\-1\end{bmatrix}$$

The second mode dies because $|\lambda_1|$ is strictly largest, and $c_1\neq0$ keeps the limit nonzero. The direction settles; $x_k$ itself keeps growing.

No. $\det(A-\lambda I)=\lambda^{2}-\tfrac{7}{10}\lambda-\tfrac{3}{10}$; times $10$, $10\lambda^{2}-7\lambda-3=(\lambda-1)(10\lambda+3)$, so $\lambda_1=1$ and $\lambda_2=-\tfrac{3}{10}$. They are distinct, so $A$ is diagonalizable, and $x_k\to0$ for every $x_0$ exactly when $\rho(A)<1$. Here $\rho(A)=1$, so $\rho(A)<1$ fails, and only just: the mode at $\lambda_1=1$ is multiplied by $1$ at every step and never dies.

One start that never decays: $x_0=(1,1)^{\mathsf T}$ gives $Ax_0=x_0$, so $x_k=x_0$ for every $k$.

From $SV=V\Lambda$, $S$ sends column $j$ of $V$ to $\lambda_j$ times itself, so the columns of the invertible $V$ would be a basis of eigenvectors of $S$. There is none: $\det(S-\lambda I)=(2-\lambda)^{2}$, and $S-2I=\begin{bmatrix}0&3\\0&0\end{bmatrix}$ forces the second entry to be $0$, so every eigenvector lies on the one line $\operatorname{span}\{(1,0)^{\mathsf T}\}$. No such $V$ exists, and $S$ is called defective.

The repeated root is not the fault: $2I$ repeats the eigenvalue $2$ and is already diagonal. What $S$ lacks is a second independent eigenvector.