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The question states: Suppose $A\in M_n(\mathbb{R})$ with rank A=r. Let V be the vectorpace of all real $n\times n$ matrices $X$ such that $AX=0$ then what is the dimension of V?

(a) r

(b)nr

(c)$n^2r$

(d) $n^2-nr$

I defined it as linear map $T:M_n(\mathbb{R})\to M_n(\mathbb{R})$ where $T(X)=AX$, So now we need to obtain nullity of the T.

Clearly option c not possible as the rank cannot be bigger than $n^2$.

They have not mentioned anything about $r$ if it is equal to $n$ then it is invertible matrix, then what can be said about $AX$?

I think we need to apply the fact that dimension of $A=r$, but how to use it I am not getting...

Thanks in advance!

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    $\begingroup$ The co-domain for $T$ is wrong. You should be able to figure out the $r=n$ case since $T^{-1}$ exists which allows you to eliminate all but one option. $\endgroup$ Commented May 2 at 16:43
  • $\begingroup$ thanks for correcting me, but why will $T^-1$ exists? $\endgroup$ Commented May 2 at 16:56
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    $\begingroup$ @mathstudent rank-nulllity. If there is indeed a correct answer among those shown, that's actually enough to answer the question. $\endgroup$ Commented May 2 at 17:06

2 Answers 2

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Note that, $T$ acting on $X$, that is $A$ being multiplied to $X$ can be seen as $A$ acting on each column vector of $X$, and the answer getting stored in the columns of $AX$.

So, if $AX$ is $0$ means that each of the column of $X$ should belong to the nullspace of $A$.

Now, since $rank(A) = r \implies ker(A) = n-r$. So, that means the dimension of vectors space from where columns of X should come from is $n-r$.

Now, we need to think about the dimension of collection of matrices whose $n$ columns come from $ker(A)$. That is equivalent to the vector space $\mathbb{R}^{n \times (n-r)}$. Thus the answer is D.

Please upvote and accept if the answer solved your query. If you need further clarification, do ask. Have a nice day.

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  • $\begingroup$ That is equivalent to $\Bbb R^{n\times (n-r)}$ $\endgroup$ Commented May 2 at 17:20
  • $\begingroup$ Kindly explain this point $\endgroup$ Commented May 2 at 17:21
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    $\begingroup$ The matrix is equivalent to choosing a vector from $\ker(A) \times ker(A) \times \ldots ker(A)$ . $ker(A)$ appears $n$ times. So the dimension of the above set is $dim(kerA) \times n$. $\endgroup$
    – Debu
    Commented May 2 at 17:26
  • $\begingroup$ let me know if you have further doubts. do accept the answer if it solved the doubt. $\endgroup$
    – Debu
    Commented May 2 at 17:30
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    $\begingroup$ yes. you are right. Infact, in general when we multiply two matrices, say $CD$, $C$ just acts on each of the column. So $CD$ is actually $[CD_{1}, CD_{2},...CD_{n}]$, where $D_{i}$ are the $i$th column of the matrix $D$. Keep this notion in mind, a lot of problems can be tackled easily with the help of this point of view. This is very easy to check, just notice the matrix multiplication carefully. Try out few examples, and this will appear natural to you. $\endgroup$
    – Debu
    Commented May 2 at 17:47
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Proposition:Let $F$ be any field and $T:F^{n\times p}\rightarrow F^{m\times p}$ is a linear transformation defined as $T(X)=AX$,where $A$ is a fixed $m\times n$ matrix.
Then for $B_1=\{E_{11},E_{21},E_{31},...,E_{n1},E_{12},E_{22},E_{32},...,E_{n2},....,E_{1p},E{2p},...,E_{np}\}$ &
$B_2=\{E_{11},E_{21},E_{31},...,E_{m1},E_{12},E_{22},E_{32},...,E_{m2},....,E_{1p},E{2p},...,E_{mp}\}$,where $E_{ij}$ are standard basis elements of $F^{n\times p}$ & $F^{m\times p}$. $$[T]^{B_2}_{B_1}=\begin{bmatrix}A & 0 & \cdots & 0 \\ 0 & A & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & A \end{bmatrix}.$$
Where matrix '$A$' appears '$p$'-times in the diagonal of $[T]^{B_2}_{B_1}$.


In view of the above proposition following results can be stated:

$1)$. $rank(T)=p\times {rank(A)}$.
$2).$ $nullity(T)=p\times{nullity(A)}$

If m=n

$1)$ $det(T)={(det(A))}^p$
$2)$ $C_T(x)=((C_A(x))^p$

Similarly you can find trace,minimal polynomial,diagonalizability e.t.c. of $T$.


Now in your case $T$ is defined from $\Bbb R^{n\times n}$ to $\Bbb R^{n\times n}$.
So we have $m=p=n$.Now rank of $A$ is given $r$,so nullity of $A$ is $n-r$ and hence by above proposition,

$nulity(T)=n\times (n-r)$.

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