Show that $\dim V = \dim W$

Question

$V$ and $W$ are subspaces of $\mathbb{C}^m$ and $\mathbb{C}^n$ respectively. $m, n$ be positive integers. Let $A$ be a $(n,m)$ matrix and $B$ be a $(m,n)$ matrix assume that $\lambda \neq 0$

$V= \{x\in \mathbb{C}^m |$ for a positive integer $k$ , $(BA-\lambda I_m)^k x=0$ holds}
$W= \{y\in \mathbb{C}^n |$ for a positive integer $l$ , $(AB-\lambda I_n)^l y=0$ holds}

Show that $\dim V= \dim W$

I have proved $BA$ and $AB$ have the same eigenvalue $\lambda$ but then how do we involve these statements, any hints would be appreciated.

You need to add a space after your \dim so the title renders correctly :) — talbi, Feb 22 '21 at 08:18
Are you aware of generalised eigenvectors and the Jordan normal form? If so, this, especially the second answer, may be useful: https://math.stackexchange.com/questions/1727590/ab-and-ba-have-identical-nonsingular-jordan-blocks — johnny10, Feb 22 '21 at 08:39
@johnny10 We can prove that $AB$ and $BA$ have the same eigenvalues but they are order or size is different $AB_{n\times n}$ and $BA_{m\times m}$, so the Jordan normal form of them should be different, am I right? Does that mean, they have the same eigenvalues but multiplicities are different? Then generalized eigenvectors are also different, am I right? — Priya, Feb 24 '21 at 01:06
Yes, the Jordan form will be different. The point is that the non-singular blocks are the same and therefore the dimensions of the generalised eigenspaces (your spaces $V$ and $W$) are the same. — johnny10, Feb 24 '21 at 09:33
@johnny10 How do we prove this from the given data? could you please give me an idea — Priya, Feb 25 '21 at 01:10
It is proved in the post I linked or now also in mathmath's answer. — johnny10, Feb 25 '21 at 11:12

mathmathmath · Accepted Answer · 2021-02-25T21:53:44.127

3

I'm editing the post after johnny10 pointing out my mistake in the previous post.

If we denote for fix $k$:

$V_k= \{x\in \mathbb{C}^m | (BA-\lambda I_m)^k x=0 \}$ and $W_k= \{x\in \mathbb{C}^n | (AB-\lambda I_n)^k x=0 \}$ then $\dim V_k =\dim W_k$.

For the proof, I just extend the answer that johnny10 gave above (but the comment doesn't allow me to type too many things so I make a post here):

The following is the same as that given by Christiaan Hattingh in the post with adapted size:

From $\begin{bmatrix} I_n & -A \\ 0 & I_m\end{bmatrix}\begin{bmatrix} AB & 0 \\ B & 0\end{bmatrix}\begin{bmatrix} I_n & A \\ 0 & I_m\end{bmatrix} = \begin{bmatrix} 0 & 0 \\ B & BA\end{bmatrix}$, we see that $\begin{bmatrix} AB & 0 \\ B & 0\end{bmatrix} \text{ and } \begin{bmatrix} 0 & 0 \\ B & BA\end{bmatrix}$ are similar so they have same Jordan form.

In particular, $\text{rank}\left (\begin{bmatrix} AB & 0 \\ B & 0\end{bmatrix} - \lambda I_{m+n}\right )^k = \text{rank}\left (\begin{bmatrix} 0 & 0 \\ B & BA\end{bmatrix} - \lambda I_{m+n}\right )^k$ for all $k$.

On the other hand, we have $\text{rank}\left (\begin{bmatrix} AB & 0 \\ B & 0\end{bmatrix} - \lambda I_{m+n}\right )^k = \text{rank}\left (\begin{bmatrix} (AB-\lambda I_n)^k & 0 \\ D & (-1)^k \lambda^k I_m\end{bmatrix} \right ) = \text{rank}(AB - \lambda I_n)^k +m$ and $\text{rank}\left (\begin{bmatrix} 0 & 0 \\ B & BA\end{bmatrix} - \lambda I_{m+n}\right )^k = \text{rank}\left (\begin{bmatrix} (-1)^k \lambda^k I_m & 0 \\ D' & (-1)^k (BA-\lambda I_m)^k\end{bmatrix} \right ) = \text{rank}(BA - \lambda I_m)^k +n$

Therefore, $n-\text{rank}(AB - \lambda I_n)^k=m-\text{rank}(BA - \lambda I_m)^k$ for all $k$ and it implies the statement above.

Hence, $\dim V =\dim W $ since in general $V$ and $W$ are just $V_k$ and $W_k$ for $k$ large (we can take $k$ to be the algebraic multiplicities).

edited Feb 25 '21 at 21:53

answered Feb 25 '21 at 05:20

mathmathmath

1,433

Keeping in mind that $\lambda\not=0$ resolves your initial doubts: If we allowed $\lambda=0$ the spaces would indeed not be of the same dimension. We however only consider generalised eigenspaces corresponding to nonzero eigenvalues. You also used $\lambda\not=0$ in your proof when simplifying the rank of the block matrix with $(-1)^k\lambda^kI_m$ in the lower right position. – johnny10 Feb 25 '21 at 11:10
Ah I see. I totally forgot we may also have eigenvalue $0$. Thanks for pointing that! I will edit the answer! – mathmathmath Feb 25 '21 at 21:50
@mathmath and johnny10 thank you for the answer, I do not understand why we fix $k$ at the beginning and the $l$ is not using on the answer, could you please tell me the very first thinking for this problem – Priya Feb 26 '21 at 00:38
@mathmath After we proved that $\begin{pmatrix}AB& 0\B&0\end{pmatrix}$ and $\begin{pmatrix}0& 0\B&BA\end{pmatrix}$ are similar how do we say "In particular $rank(..)^k= rank(..)^k$ " for all $k$ ? – Priya Feb 26 '21 at 01:42
And what is $D$ and $D^\prime$? – Priya Feb 26 '21 at 01:49
1

Priya, for your first question: V (and W) is just the union of all $V_k$ (or $W_k$), and in fact, they will be equal to $V_k$ (or $W_k$) for some $k$ large enough (you should consult the theory of generalized eigenspaces for this point) – mathmathmath Feb 28 '21 at 06:22
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For the second question, please look at the formula to compute the number of Jordan blocks for a matrix. The rank^k will be there. – mathmathmath Feb 28 '21 at 06:23
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For the last question, they are just some matrices of suitable size (because when we do this $ (\begin{pmatrix} AB & 0 \ B & 0 \end{pmatrix} - \lambda I_{m+n} \right )^k $ we will still get a triangular matrix) – mathmathmath Feb 28 '21 at 06:24

score 1 · Answer 2 · answered Mar 03 '21 at 18:50

As mathmath has already posted a handmade solution, I feel free to write a more sophisticated proof.

Recall that if $T$ is an endomorphism and $\sigma$ is an eigenvalue, its geometric multiplicity is the dimension of $\ker (T-\sigma I)$, while its algebraic multiplicity is the multiplicity of $\sigma$ a a root of the characteristic polynomial of $T$, $p_T(x)$ and you know that the first one is less than or equal to the second one.

When studying the Jordan normal form, one needs to consider the generalised eigenspaces $\ker (T-\sigma)^k$ and they form a tower that stops at some point $$\ker (T-\sigma I)^1 \subsetneq \ker (T-\sigma I)^2 \subsetneq \ldots \subsetneq \ker(T-\sigma I)^N = \ker(T-\sigma I)^{N+1}=\ldots$$And it turns out that the algebraic dimension of $\sigma$ is the dimension of this top space $\ker (T-\sigma I)^N$.

Now, turning back to your problem, and using your notation, we have $$V= \ker (BA-\lambda I) \oplus \ker(BA-\lambda I)^2 \oplus \ldots \qquad W= \ker (AB-\lambda I) \oplus \ker(AB-\lambda I)^2 \oplus \ldots$$Therefore, $\dim V$ is the algebraic multiplicity of $\lambda$ as an eigenvalue of $BA$ and $\dim W$ is the algebraic multiplicity of $\lambda$ as an eigenvalue of $AB$.

It is well known that if $A$ and $B$ are square matrices of the same dimension, then the characteristic polynomials of $AB$ and $BA$ are the same (I'll let you think about this or look it by yourself). What is not so well known is that a similar result holds when $A$ and $B$ are not square, as in your case. If $m<n$, for example, add zeroes at the matrices $A$ and $B$ to get two square $n \times n$ matrices $A'$, $B'$, and use that $p_{A'B'}(x) = p_{B'A'}(x)$ to obtain $p_{AB}(t) = t^{n-m}p_{BA}(t)$ and from there and all what I have commented, the result follows. Note the importance that $\lambda \neq 0$

Thank you very much for the answer. As you said in this case $A$ and $B$ are not square so algebraic multiplicity of $\lambda$ of $AB$ and alg. mul of $\lambda$ of $BA$ are not the same. Therefore the above direct sum expression of $V$ and $W$ are not equal. So how can we figure this out for this case — Priya, Mar 05 '21 at 04:22
How do we compare the given $l$ and $k$ with $N$ in your answer — Priya, Mar 05 '21 at 04:25
Hi Aitor, thanks for the nice argument with observation $p_{AB}(t) = t^{n-m}p_{BA}(t)$! I'm very glad I learn this nice, quick trick now. However, I don't think we can write $V,W$ as direct sums using Priya notation. — mathmathmath, Mar 06 '21 at 19:21
Priya, using $p_{AB}(t) = t^{n-m}p_{BA}(t)$ in Aitor comment, the algebraic multiplicities of $\lambda$ of both matrices are the same ($N$). In your definition, $k,l$ are not given. For example, the definition of $V$ says: $x\in V$ if there is some number $k$ such that $(BA-\lambda I_m)^k x=0$. (So clearly if this is the case then $(BA-\lambda I_m)^t x=0$ for all $t\ge k$.
Therefore, as in my post and using the $N$ in Aitor post, $V=\ker (BA-\lambda I)^N$. — mathmathmath, Mar 06 '21 at 19:26
mathmath- Since there is no $\lambda=0$, $n-m$ term of $p_{AB}(t)=t^{n-m}p_{BA}(t)$ should be equal zero then $n=m$. Therefore the algebraic multiplicity of $\lambda$ of both matrices are the same. Am I right? — Priya, Mar 09 '21 at 11:27

Show that $\dim V = \dim W$

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