# $A$ be a $n \times n$ matrix with $\text{rank}\,(A)\lt n$: multiple choice question

Let $A$ be an $n \times n$ real non-zero matrix of rank less than $n$. Then one of the following is true? :

(A) there exists an $n \times n$ real non-zero matrix $B$ such that $BA = 0$.

(B) there may not always exist an $n \times n$ real non-zero matrix $B$ such that $BA = 0$.

(C) there exists an $n \times n$ real non-zero matrix $B$ such that $BA = I$.

(D) if $B$ is such that $BA = 0$, then $AB = 0$.

My Attempt: Consider a $2\times 2$ matrix $A$ of the form $\begin{pmatrix} 0 &1 \\ 0 & 0 \end{pmatrix}$

and a $2\times 2$ matrix $B$ of the form $\begin{pmatrix} 0 &a \\ 0 & c \end{pmatrix},$ where $a,c$ are any non zero real numbers.

Then I have that $BA=0$ holds. So option $(A)$ may hold.

Option $(B)$ can also hold if we take $2\times 2$ matrix $B$ of the form $\begin{pmatrix} 1 & a \\ 2 & c \end{pmatrix}$

$(C)$ is clearly false.

Also $(D)$ is also false as is evident if we take $2 \times 2$ matrix $A$ of the form $\begin{pmatrix} 0 &1 \\ 0 & 0 \end{pmatrix}$ and $2 \times 2$ matrix $B$ of the form $\begin{pmatrix} 0 &a \\ 0 & c \end{pmatrix}$

So I am confused between $(A)$ and $(B)$. Which one should be the right choice?

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This is not multiple choice, is it? Are you supposed to mark which are true and which false? – leonbloy Jan 21 '13 at 16:43
A and B are clearly contradictory, if one is false the other is true – leonbloy Jan 21 '13 at 16:45
It is definitely a multiple choice question and i have to select only one option and ignore others. – user52976 Jan 21 '13 at 16:46
then A) is the winner. – Berci Jan 21 '13 at 16:57
@Berci thanks a lot sir. – user52976 Jan 21 '13 at 16:58

Showing existence of a matrix $B$ such that $AB = 0$ requires only one such matrix (whatever the value of $n$.)

You shown such existence (option (A)) for $n = 2$.

Hint: Can you generalize to consider other $n\times n$ matrices A whose rank is less than n? If yes, that is, if you can show existence of "even one" matrix $B$ such that $BA = 0$ whatever the value of $n$, then (A) is the option to choose.

Note that since the rank of $A$ is less than $n$, (by hypothesis), the columnspace given by $\text{im} A=\{Ax:x\in\Bbb R^n\}$ is therefore not of full dimension.

So choose any matrix $B$ which maps $\text{im} A$ to zero but sends a vector out from $\text{im} A$ to some non-zero vector.

Such a $B$ can certainly be constructed for any $n\times n$ matrix $A$ with rank less than $n$.

So we have generalized to show existence of a $B$ such that $BA = 0$ whatever the dimension $n$.

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Magic A, winner A. – S. Snape Jan 21 '13 at 17:33

If rank of $A$ is less than $n$, then the columnspace ($im A=\{Ax:x\in\Bbb R^n\}$) is not full dimension, so any $B$ which maps $im A$ to zero but sends a vector $v\notin im A$ to a nonzero vector, will do the job.

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