Does Nullspace(A) = Nullspace(C) imply Nullspace(AB) = Nullspace(CB)

linear algebramatrices

I recently thought of the following linear algebra statement:

For arbitrary matrices $\mathbf{A}, \mathbf{B}, \mathbf{C}$, we have
$ \mathcal{N}(A) = \mathcal{N}(C) \implies \mathcal{N}(AB) = \mathcal{N}(CB) $

Is this statement correct?

I came up with the following proof attempt:

Assume $\mathcal{N}(A) = \mathcal{N}(C)$. Then,
\begin{equation} \label{eq1}
\begin{split}
x\in\mathcal{N}(AB) & \iff x\in\left(\mathcal{N}(B) + \left(\mathcal{N}(A)\cap\mathcal{C}(B)\right)\right) \\
& \iff x\in\left(\mathcal{N}(B) + \left(\mathcal{N}(C)\cap\mathcal{C}(B)\right)\right) \\
& \iff x\in\mathcal{N}(CB).
\end{split}
\end{equation}

Is this proof correct?

(I denote nullspaces and column spaces with $\mathcal{N}$ and $\mathcal{C}$, respectively.)

Best Answer

It’s not true that $\mathcal{N}(AB) = \mathcal{N}(B) + \bigl(\mathcal{N}(A) \cap \mathcal{C}(B)\bigr)$. For one, if the matrices aren’t square, then these vector subspace operations might not be able to be performed due to differing underlying spaces.

A more appropriate equivalence is $x \in \mathcal{N}(AB) \iff Bx \in \mathcal{N}(A)$. From which it follows that

$$\begin{align} x \in \mathcal{N}(AB) &\iff Bx \in \mathcal{N}(A) \\ &\iff Bx \in \mathcal{N}(C) \\ &\iff x \in \mathcal{N}(CB), \end{align} $$

whenever $\mathcal{N}(A) = \mathcal{N}(C)$.

Related Solutions

[Math] $A, A^T$: Same left nullspace. What if row space = column space? [GStrang P193 3.6.25]

Only posterior to seeing the answers do I realise that (b) and (c) are false. Anterior to finding counterexamples, what alludes/suggests that (b) and (c) are false?

You've put phrasing like this into many of your recent questions (often with different grandiloquent word choices for "before" and "after"). But it seems a little weird to me to make such a point of asking how one would know how to solve the problem in advance: if you know that a statement is false "anterior" to finding counterexamples, then you don't need to do the problem at all. The whole point of solving problems is that you've learned something "posterior" to solving the problem that you didn't know "anterior" to working on it.

Also, how were the counterexamples devised?

There are many ways of finding counterexamples for questions like these. If you want to know the mental processes of the person who wrote up the solution, ask them.

Maybe you are asking "How would I usefully get started thinking about questions like this?" If you're asked whether $X$ is always true, then (assuming you don't know the answer already...) you should start by looking at examples where $X$ can be either true or false. If in any of your examples you find that $X$ is not true, you've answered the question. If you keep trying examples and $X$ is always true, you need to either look for a common feature of the examples that makes $X$ hold, or you need to switch to a broader class of examples. This process is dynamic: you could do this several times before you finally come up with an answer, and what you write as your answer will probably not record your entire thought process but merely what worked in the end.

Your starting point is a good one as far as it goes: the "left null spaces of $A$ and $A^T$" are just the null spaces of $A^T$ and $A$, so the question is slightly strangely stated: it might as well ask whether the null space of $A$ must be equal to the null space of $A^T$. Is this true? As the answer points out, it doesn't make sense if the matrix is not square, so that's a kind of answer, but now let's restrict to square matrices. We try some examples, startin with the simplest: e.g. the zero matrix. Then both null spaces are all of $\mathbb{R}^n$. Or the identity matrix. Then both null spaces are $0$. How about $A = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right]$. The null space of $A$ is spanned by $e_2$. What about the null space of $A^T$? Oh, of course it's the same as the null space of $A$ because $A^T = A$. So we see that in order to find counterexamples we need to look past symmetric matrices. What's the simplest nonsymmetric matrix? Maybe $\left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right]$. Here the null space of the matrix is spanned by $e_1$ and the null space of the transpose is spanned by $e_2$. They are not equal, so we've answered the question.

Let's look at part c). The question is whether equality of row and column spaces forces the matrix to be symmetric. Again if I just play around with nonsymmetric matrices then I'll soon find $\left[ \begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array} \right]$. It's a counterexample because it's not symmetric and because its row and column space are equal. In fact, they're equal because they're both $\mathbb{R}^2$. Since the row space of an $n \times n$ matrix is $\mathbb{R}^n$ if and only if the matrix is invertible if and only if the column space is $\mathbb{R}^n$, we see that we've found a much broader class of counterexamples, so perhaps we record that as an answer rather than the special case we found just by exploring the example space.

Find a vector in the column space of both matrices

When finding common vectors in the column spaces of the matrices $A$ and $B$ (since in this case you only have two vectors(columns) per matrix you know they are not linearly dependent because they are not multiples of one another)

you are looking for common vectors in the spans of the form $a_1 \begin{bmatrix} 1\\1\\1\end{bmatrix}+a_2\begin{bmatrix} 2\\3\\2\end{bmatrix} = A\begin{bmatrix} a_1\\a_2\end{bmatrix} = A\vec{x} = b_1\begin{bmatrix} 5\\6\\5\end{bmatrix} + b_2\begin{bmatrix} 4\\3\\1\end{bmatrix} = B\begin{bmatrix} b_1\\b_2\end{bmatrix}=B\vec{y}$

(can a linear combination of the column vectors of $A$ equal a linear combination of the column vectors of $B$, i.e. the spaces share a common vector)

and $[\: A \:\: B \:] \begin{bmatrix} \vec{x}\\ -\vec{y}\end{bmatrix} = \vec{0}$ means essentially the same thing :=

$\begin{bmatrix} 1&2&5&4\\1&3&6&3\\1&2&5&1\end{bmatrix}\begin{bmatrix} a_1\\a_2\\-b_1\\-b_2\end{bmatrix} = a_1 \begin{bmatrix} 1\\1\\1\end{bmatrix}+a_2\begin{bmatrix} 2\\3\\2\end{bmatrix} - b_1\begin{bmatrix} 5\\6\\5\end{bmatrix} - b_2\begin{bmatrix} 4\\3\\1\end{bmatrix} = A\vec{x} -B\vec{y}=\vec{0}$

Since you are in $3$-space and you have two linearly independent vectors for each column space of $A$ and $B$, the subspaces generated by the span of those column vectors will be planes (if you had just one vector, the subspace would be a line, three linearly independent vectors, all of $3$-space), and as mentioned by the textbook answer, there are exists at least one nonzero solution (every homogeneous set of equations is consistent, has the zero solution) to the homogeneous set of equations

$$[\: A \:\: B \:] \begin{bmatrix} \vec{x}\\ -\vec{y}\end{bmatrix} = \vec{0} $$

because there are more unknowns than equations. So the planes either intersect in a line, or completely overlap.

This means the column spaces are not orthogonal by the definition of orthogonal because since they intersect, you can find a nonzero vector $\vec{v}$ s.t. $\vec{v}\in \text{col}(A),\text{col}(B)$ and $\vec{v} \cdot \vec{v} = ||\vec{v}||^2 \not= 0$.

Best Answer

Related Solutions

[Math] $A, A^T$: Same left nullspace. What if row space = column space? [GStrang P193 3.6.25]

Find a vector in the column space of both matrices

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