Good true-false linear algebra questions? Can you suggest me a collection of true-false linear algebra questions, like the ones found in the MIT exams, if possible with solutions (i.e. explanations)?
Sorry if it turns out that my request is off-topic, but if it is, please suggest where to ask it properly
 A: A question I am fond of asking my students: Does the vector space of polynomials of degree $\leq n$ have a basis consisting of polynomials with same degree?
Answer: yes. 
A: If $T: \mathbb{R}^3 \to \mathbb{R}$ is a linear transformation, then $T(x,y,z) = ax + by + cz$ for some $a,b,c \in \mathbb{R}$.
If $S,T: V \to V$ are linear transformations and $v \in \ker T$, then $v \in \ker S \circ T$.
If two non-zero vectors are linearly independent, then one must be a scalar multiple of the other.
If $T$ is invertible, then $\ker T = \{\textbf{0}\}$.
If $\{v_1, v_2, \dots, v_n\}$ are linearly independent in $\mathbb{Q}^n$, then they must form a basis for $\mathbb{Q}^n$.
There can never be a linear map between finite dimensional spaces of different dimension. 
A: Don't know how difficult these need to be. Here are some rather simple ones:


*

*For every linear map $f : \mathbb R^m \to \mathbb R^n$ there is a unique $m\times n$ matrix with $f(v) = Av$ for all $v\in \mathbb R^m$.
Answer: No, its actually an $n\times m$ matrix. The expression $Av$ would not even make sense.

*If $AB = AC$ and $A\neq 0$, then $B = C$ for all $n\times n$ matrices $A,B,C$. Answer: No, take $n = 2$, $A = B = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and $C = 0$.

*The determinant of a matrix with integer entries is an integer. Answer: Yes, take a look at (for example) the Leibniz formula for determinants.

*$\det : \mathbb{R}^{n\times n} \to \mathbb{R}$ is a linear map for all $n$. Answer: No, unless $n = 1$. Instead we have $\det(cA) = c^n\det(A)$ for $c\in \mathbb R$ and no "obvious rule" to compute $\det(A+B)$.

*We have $(A+B)^n = \sum_{i=1}^n \binom{n}{i} A^{n-i} B^i$ for all $n\times n$ matrices $A,B$. Answer: No. It is true, if $AB = BA$.

*If $m,n \geq 0, A \in \mathbb R ^{m\times n}, B\in \mathbb R^ {n\times m}$ with $AB = I_m$ and $BA = I_n$ then $m=n$. Answer: Yes, this can be seen using the dimension formula / rank-nullity theorem.

A: Here are some random ones that have a narrower scope: they concern more about matrices than about linear algebra in general.


*

*If a system of homogeneous linear equations is overdetermined (i.e. it has more equations than variables), it must have only the trivial solution. (False. Consider $Ax=0$, where $A$ has linearly dependent columns.)

*If $A$ and $B$ are two matrices such that $AB=I$, then $BA=I$. (False. If $A,B$ are rectangular matrices and $AB=I$, then $BA$ must have deficient rank.)

*Let $A,B$ be two rectangular matrices and $C=AB^\top$. If $C^3-5C+I=0$, then $A$ has linearly independent rows. (True. $C$ is invertible and $A$ has full row rank.)

*There exist four vectors $u,v,w,x\in\mathbb R^4$ such that $uv^\top+vw^\top+wx^\top+xu^T=I_4$.
(False. Let $P$ be the augmented matrix $(u,v,w,x)$ and
$$
C=\pmatrix{0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0}.
$$
Then $uv^\top+vw^\top+wx^\top+xu^T=PCP^T$ and its determinant is $-\det(P)^2$ that cannot be 1.)

*$vv^\top$ is diagonalisable for any complex vector $v$. (False. Consider $v=\pmatrix{1\\ i}$. Then $vv^\top$ is nonzero but nilpotent -- its square is zero, hence non-diagonalisable.)

*If $P$ is an invertible matrix of the same size as $A$, then $\operatorname{adj}(PAP^{-1})=P\operatorname{adj}(A)P^{-1}$. (True, because $\operatorname{adj}(M)$ is a polynomial in $M$.)

*Let $A,B\in M_3(\mathbb R)$ and $ABA=A^2$. In each of the following cases, is it true that $\operatorname{adj}(AB)=\operatorname{adj}(A)B$?


*

*$A$ is nonsingular.

*$A$ has rank 2.

*$A$ has rank 1.


(True, false, true.) The first case is trivial because $B=I$. In the second case, consider the case where $A=\operatorname{diag}(1,1,0)$ and $B-A$ is a matrix whose only nonzero row is the last one. In the third case, the rank deficiencies in both $A$ and $AB$ are at least 2. Hence their adjugate matrices are zero.

*Let $\varepsilon>0$ and $p$ be a polynomial with real coefficients such that $p(A)=0$ for every real square matrix $A$ that is entrywise smaller than $\varepsilon$. Then $p=0$. (True. As $p(tI)=0$ when $t$ is small, we get $\operatorname{diag}(p(t),\ldots,p(t))=0$ when $t$ is small. Hence $p$ has infinitely many roots.)

*Let $V$ be the vector space of all cubic polynomials with real coefficients. Let $T$ be the linear operator defined by $T(p)(x)=p'(x)$. Then its adjugate is $\operatorname{adj}(T)=T^3$. (False. The characteristic polynomial of $T$ is $x^4$. Hence $\operatorname{adj}(T)=-T^3$.)

A: A simple but not trivial question:
(T/F): If $n\times n$ matrix $A$ is of rank $n$, then $A$ is diagonalizable by a similarity transformation $$ D = P^{-1}AP$$
Another, slightly harder one:  
If $P(x)$ is a polynomial and $A$ is a matrix with eigenvalues $\lambda_1 \ldots \lambda_n$, then $P(A)$ has eigenvalues $P(\lambda_1) \ldots P(\lambda_n)$
