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$A$ is defined as an $m\times m$ matrix which is not invertible. How can i show that there is an $m\times m$ matrix $B$ where $AB = 0$ but $B$ is not equal to $0$?

For the solution of this question I think giving an example is not enough because it is too easy to solve this by giving an example, so how can I show that $B$ is not the $0$ matrix?

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Hint: $AB=0$ if and only if every column of $B$ is in the null space of $A$. – Henning Makholm Nov 23 '12 at 13:16

The fact that $A$ is not invertible implies that $A$ has nontrivial kernel, i.e. there exists a vector $v$ such that $Av=0$. Now let $B=vv^T$.

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The OP didn't say that he is working over $\mathbb{R}$. :) – darij grinberg Aug 16 '14 at 12:18
@darijgrinberg: did I? – Martin Argerami Aug 16 '14 at 16:11
Oops, sorry, you didn't. Good answer! – darij grinberg Aug 16 '14 at 16:34

Here is a nice, though probably not practical, way to find $B$.

The matrices $I=A^0$, $A=A^1$, $A^2$, $\ldots$, $A^{m^2}$ must be linearly dependent because the space of $m\times m$ matrices has dimension $m^2$. This means that $A$ is a zero of some polynomial of degree at most $m^2$.

Let $n$ be the smallest degree of a polynomial for which $A$ is a zero. Then we have $\alpha_0 I + \alpha_1 A + \cdots + \alpha_n A^n=0$ and so $A(\alpha_n A^{n-1} + \cdots + \alpha_1 I) = -\alpha_0 I$. Let $B=\alpha_n A^{n-1} + \cdots + \alpha_1 I$. If $\alpha_0\ne0$ then $(-\alpha_0)^{-1}B$ is an inverse for $A$. So, if $A$ is not invertible, then $\alpha_0=0$ and $AB=0$. Note that $B\ne0$ because $n$ is minimal.

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Very nice!$\ \ \ $ – Martin Argerami Nov 24 '12 at 16:38

How can I show that [if an $m\times m$ matrix $A$ is not invertible,] there is an $m\times m$ matrix $B$ where $AB=0$ but $B$ is not equal to $0$?

I am including the following, in case you have not yet encountered the definitions or concepts of "null space", "rank/dimension", or "kernel":

You answered your own question in that you need only show that when $A$ is not invertible, then there is some (indeed many) non-zero matrix $B$ for which $AB = 0$. To answer this question, it suffices to show existence of a non-zero matrix $B$ such that $AB \neq 0$. That is, you thereby show that $B$ need not be identically zero.

You can prove this by induction on $m$,

  • Show the proposition is true for the base case $m = 1$.
  • Assume the proposition holds for $(m-1)\times (m-1)$ matrices $A$ and $B$.
  • Show that this assumption implies that the proposition must be true for $m\times m$ matrices, as well.
  • Conclusion: for all $m\times m$ non-invertible matrices $A$, there is an $m \times m$ matrix $B$ such that $B \neq 0$, but $AB = 0$.

You can also use the property of the determinant to get more intuition about this problem (property: $\text{det}\,(AB) = \text{det}\,(A)\,\text{det}\,(B)$):

Taking the determinant of each side of the equation $AB = 0$, and knowing that the determinant of an non-invertible matrix $A$ is equal to $0$, we have:

$$\text{det}(AB) = \text{det}(0)$$ $$\iff \text{det}A \;\text{det}B = 0\cdot \text{det} B = 0$$ for all matrices $B$. We know that for every invertible matrix $B$, $B\not \equiv 0$, and we also know that for an invertible $B$, $\text{det}B \neq 0$, but nonetheless $\text{det} A \;\text{det} B = 0\cdot \text{det} B = 0$.

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Suppose $A$ is equivalent to \begin{bmatrix} I_r & 0 \\ 0 & 0 \end{bmatrix} and $$ A=P\begin{bmatrix} I_r & 0 \\ 0 & 0 \end{bmatrix}Q $$ where $P,Q$ are invertible, and $r$ is the rank of $A$, thus $$ AB=0\iff\begin{bmatrix} I_r & 0 \\ 0 & 0 \end{bmatrix}QB=0 $$ Suppose $$ QB= \begin{bmatrix} C_1 \\ C_2 \end{bmatrix} $$ where $C_1$ has $r$ rows, we have $C_1=0$, so $$ B=Q^{-1}\begin{bmatrix} 0 \\ C_2 \end{bmatrix} $$ where $C_2$ is an arbitrary $(m-r)\times m$ matrix.

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Since $A$ is not invertible, by the Invertible Matrix Theorem there is a nontrivial solution $x_0$ of the homogeneous equation $Ax_0 = 0$. Take $x_0$ as the first column of $B$ and take the remaining columns of $B$ to be zero. That gives $AB=0$ but $B$ is not the zero matrix.

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