M1 Practice Set

• $AB$: $(3\times5)(5\times2)$, inner dimensions meet. 3 × 2
• $BA$: $(5\times2)(3\times5)$, $2\neq3$. undefined
• $A^{\mathsf T}B$: $(5\times3)(5\times2)$, $3\neq5$. undefined
• $AB^{\mathsf T}$: $(3\times5)(2\times5)$, $5\neq2$. undefined
• $B^{\mathsf T}A^{\mathsf T}$: $(2\times5)(5\times3)$. 2 × 3 — and this is $(AB)^{\mathsf T}$, which must be $2\times3$.

(a) Linear.

(b) Not linear. The zero test: $f(0)=c\neq0$, and a linear map must send $0$ to $0$. One point is enough.

(c) Not linear. Here $f(0)=0$, so the zero test passes and is not enough. Homogeneity fails: $f(-x)=\lVert-x\rVert=\lVert x\rVert=f(x)$, while linearity demands $-f(x)$. With $x=(3,4)$: $f(-x)=5$ but $-f(x)=-5$.

(d) Not linear. $f(2x)=(4x_1x_2,\,2x_1)$ while $2f(x)=(2x_1x_2,\,2x_1)$; these differ unless $x_1x_2=0$.

$Ae_j=a_j$: asking for one elementary direction reads off one column.

$$Ce_1=\begin{bmatrix}2\\3\end{bmatrix},\qquad Ce_2=\begin{bmatrix}-1\\0\end{bmatrix},\qquad Ce_3=\begin{bmatrix}0\\4\end{bmatrix}$$

By columns — the weights come from $x$:

$$Cx=1\begin{bmatrix}2\\3\end{bmatrix}+2\begin{bmatrix}-1\\0\end{bmatrix} -1\begin{bmatrix}0\\4\end{bmatrix}=\begin{bmatrix}0\\-1\end{bmatrix}$$

By rows: $2(1)+(-1)(2)+0(-1)=0$ and $3(1)+0(2)+4(-1)=-1$. Same arithmetic, different order; the column reading is the one that tells you $Cx$ lies in $\mathcal C(C)$.

$$PQ=\begin{bmatrix}5&-4\\1&-2\end{bmatrix},\qquad (PQ)^{\mathsf T}=\begin{bmatrix}5&1\\-4&-2\end{bmatrix}$$ $$Q^{\mathsf T}P^{\mathsf T}=\begin{bmatrix}5&1\\-4&-2\end{bmatrix},\qquad P^{\mathsf T}Q^{\mathsf T}=\begin{bmatrix}3&1\\6&0\end{bmatrix}$$

$(PQ)^{\mathsf T}=Q^{\mathsf T}P^{\mathsf T}$. The transpose reverses the order, and the shapes say why: $Q^{\mathsf T}P^{\mathsf T}$ is $(p\times n)(n\times m)$, so $n$ meets $n$. Both products happen to be defined here only because the matrices are square.

Membership is a system to solve: $c_1v_1+c_2v_2=b$.

$b$ is in the span. The first coordinate forces $c_1=2$, the second forces $2(2)+c_2=3$ so $c_2=-1$, and the third checks: $2(1)+(-1)(3)=-1$.  $b=2v_1-v_2$.

$b'$ is not. The same two coordinates force the same $c=(2,-1)$, and the third coordinate is then $-1$, not $-4$. No choice of weights works.

(a) A line through the origin.

(b) Still a line — $(2,2)=2(1,1)$, so the second arrow buys nothing.

(c) A plane through the origin, the $x_1x_2$-plane.

(d) The same plane. $(1,1,0)$ is already a mixture of the first two, so the list of three spans no more than the list of two. Counting arrows is not counting dimensions.

(a) Subspace. A line through the origin; $\alpha u+\beta v$ satisfies the same equation.

(b) No. $0\notin S_2$, since $0\neq3(0)+1$. Fastest disqualifier there is.

(c) No. This one is the trap: it contains $0$ and it is closed under scaling, so two of the three checks pass. It fails closure under addition — $(1,0)$ and $(0,1)$ are both in it, $(1,1)$ is not.

(d) Subspace. A plane through the origin.

Two independent columns in $\mathbb R^3$ span a plane. A normal to it satisfies $n\cdot a_1=n\cdot a_2=0$, giving $n=(1,-1,-1)$, so

$$\mathcal C(A)=\{y\in\mathbb R^3:\ y_1-y_2-y_3=0\}$$

$b$ is reachable: $3-1-2=0$. Solving, $x=(1,1)^{\mathsf T}$, and indeed $1(1,0,1)+1(2,1,1)=(3,1,2)$.

$b'$ is not: $3-1-1=1\neq0$. It misses the plane by one, and one unit is enough — there is no $x$ at all, not merely a bad one.

Being in $\mathcal C(A)$ means being an output, so there are $x_1,x_2$ with $Ax_1=b_1$ and $Ax_2=b_2$. Then

$$\alpha b_1+\beta b_2=\alpha Ax_1+\beta Ax_2=A(\alpha x_1+\beta x_2),$$

which exhibits an input producing it. Linearity is the only thing used. Together with $A0=0$ this is the whole proof that $\mathcal C(A)$ is a subspace of $\mathbb R^m$.

The columns give it away: $a_3=a_1+a_2$, so $a_1+a_2-a_3=0$, which says

$$A\begin{bmatrix}1\\1\\-1\end{bmatrix}=0.$$

Elimination confirms there is nothing else. The reduced form is $\begin{bmatrix}1&0&1\\0&1&1\\0&0&0\end{bmatrix}$, so with $x_3=t$ free, $x_1=-t$ and $x_2=-t$:

$$\mathcal N(A)=\operatorname{span}\{(-1,-1,1)^{\mathsf T}\},\qquad\dim\mathcal N(A)=1.$$

$(1,1,-1)$ and $(-1,-1,1)$ span the same line.

$Ax_1=(1,2,3)^{\mathsf T}$ and $Ax_2=(2+1-2,\;4+1-3,\;6+2-5)^{\mathsf T} =(1,2,3)^{\mathsf T}$. Both work.

$x_2-x_1=(1,1,-1)^{\mathsf T}$, which is in $\mathcal N(A)$. It had to be:

$$Ax_1=Ax_2\ \Longrightarrow\ A(x_2-x_1)=0.$$

Two different inputs with the same output differ by a null-space vector, always.

(a) Independent.

(b) Dependent: $v_2=2v_1$, so $2v_1-v_2=0$ with weights $(2,-1)\neq0$.

(c) Dependent: $v_1+v_2-v_3=0$.

(d) Independent. Stack them as columns of $V$ and eliminate: $\mathcal N(V)=\{0\}$, so the only mix that vanishes is the zero mix.

$\dim\mathcal N(J)=7-6=1$.

Such a $z$ moves the joints while the end-effector stays perfectly still — the arm changes its own posture and the tool does not move. Formally $J(\dot q+z)=J\dot q$, so adding $z$ to any joint velocity leaves the end-effector velocity unchanged. That one free direction is the redundancy of a seven-joint arm doing a six-dimensional task, and it is what a secondary objective is allowed to use.

Non-empty: $A0=0$, so $0\in\mathcal N(A)$.

Closed under mixtures: take $u,v\in\mathcal N(A)$, so $Au=0$ and $Av=0$. For any $\alpha,\beta\in\mathbb R$,

$$A(\alpha u+\beta v)=\alpha Au+\beta Av=\alpha0+\beta0=0,$$

so $\alpha u+\beta v\in\mathcal N(A)$. Those are the two conditions, so $\mathcal N(A)$ is a subspace — and note it sits in $\mathbb R^n$, the input side, not in $\mathbb R^m$.

$$A\longrightarrow R=\begin{bmatrix}1&2&0&3\\0&0&1&1\\0&0&0&0\end{bmatrix}$$

Pivots sit in columns 1 and 3. So $x_1$ and $x_3$ are pivot variables, and $x_2$ and $x_4$ are free — two of them.

A basis for $\mathcal C(A)$ is the pivot columns of $A$ — never the columns of $R$, whose column space is a different set:

$$\mathcal C(A)=\operatorname{span}\left\{(1,2,1)^{\mathsf T},\,(0,1,1)^{\mathsf T}\right\}$$

For $\mathcal N(A)$, set one free variable to 1 and the other to 0 in turn:

$$\mathcal N(A)=\operatorname{span}\left\{(-2,1,0,0)^{\mathsf T},\, (-3,0,-1,1)^{\mathsf T}\right\}$$

$\operatorname{rank}A=2$ (two pivots), $\operatorname{nullity}A=2$ (two free variables), and $2+2=4=n$. Rank–nullity counts the input side: every column is either a pivot or free, and there are $n$ of them.

Solve $c_1(1,1)+c_2(1,-1)=(4,2)$: adding the two equations gives $2c_1=6$, so $c_1=3$ and then $c_2=1$.

$$[x]_B=\begin{bmatrix}3\\1\end{bmatrix},\qquad 3\begin{bmatrix}1\\1\end{bmatrix}+1\begin{bmatrix}1\\-1\end{bmatrix} =\begin{bmatrix}4\\2\end{bmatrix}$$

The weights are unique — that is what a basis buys, and it is why coordinates are well defined at all.

• $\dim\mathcal C(B)=r=6$, a 6-dimensional subspace of $\mathbb R^7$.
• $\dim\mathcal N(B)=n-r=10-6=4$.
• Four free variables.
• No. Uniqueness needs $\mathcal N(B)=\{0\}$. Here the null space is four-dimensional, so any solution comes with a four-parameter family of companions. The count is: no solutions, or infinitely many — never exactly one.

Note also $6<7$, so $\mathcal C(B)\neq\mathbb R^7$ and some $b$ are unreachable.

$r=n$, full column rank. The columns are independent and $\mathcal N(A)=\{0\}$, so $Ax=b$ has at most one solution — uniqueness, with no promise that any solution exists.

$r=m$, full row rank. $\mathcal C(A)=\mathbb R^m$, so $Ax=b$ has at least one solution for every $b$ — existence, with no promise it is unique.

Both at once forces $m=n=r$, which is exactly invertibility.

Reduce the augmented matrix:

$$\left[\begin{array}{ccc|c}1&1&2&1\\2&1&3&2\\3&2&5&3\end{array}\right] \longrightarrow \left[\begin{array}{ccc|c}1&0&1&1\\0&1&1&0\\0&0&0&0\end{array}\right]$$

The bottom row is $0=0$, so the system is consistent. With $x_3=t$ free, $x_1=1-t$ and $x_2=-t$:

$$x=\begin{bmatrix}1\\0\\0\end{bmatrix}+t\begin{bmatrix}-1\\-1\\1\end{bmatrix}, \qquad t\in\mathbb R$$

One particular solution plus the null space — a line in $\mathbb R^3$, not through the origin.

(a) Infinitely many — a three-parameter family, since $\dim\mathcal N(A)=6-3=3$ and the solution set is $x_p+\mathcal N(A)$.

(b) None. Consistency is exactly membership: $Ax=b$ is solvable if and only if $b\in\mathcal C(A)$.

Existence is decided on the output side, uniqueness on the input side. They are separate questions and this matrix answers them differently.

$A0=0\neq b$, so $0\notin\mathcal S(A,b)$ and the fastest disqualifier applies at once.

Closure fails too: if $Ax_1=b$ and $Ax_2=b$ then $A(x_1+x_2)=2b\neq b$.

What it is is an affine set — a subspace pushed off the origin, $x_p+\mathcal N(A)$. The direction of the set is a subspace; its position is not.

$$\mathcal S(A,b)=\left\{\begin{bmatrix}2\\0\\1\end{bmatrix} +t\begin{bmatrix}1\\-1\\0\end{bmatrix}:t\in\mathbb R\right\} =\begin{bmatrix}2+t\\-t\\1\end{bmatrix}$$

Yes, with $t=-2$: the first coordinate gives $2+t=0$, and then $-t=2$ and the third coordinate is $1$. All three agree, so $(0,2,1)$ is on the line.

Row two is twice row one, so eliminating gives $0=a-2$. The system is consistent exactly when $a=2$; for any other $a$ the last row reads $0=\text{something nonzero}$.

With $a=2$ the single surviving equation is $x_1+2x_2=1$, so

$$x=\begin{bmatrix}1\\0\end{bmatrix}+t\begin{bmatrix}-2\\1\end{bmatrix}.$$

Seen on the output side: $\mathcal C(A)$ is the line spanned by $(1,2)$, and $(1,a)$ lies on it only for $a=2$.