MHB Characteristic Equations and Commutativity

[A.Magnus]

How do I go about proving that for two matrices of same size, $M$ and $N$, that the characteristic polynomials of $MN$ and $NM$ are the same? If $M$ and $N$ are inversible, then the proof are very straightforward, for example, I can have

$|MN - \lambda I| = |MN - \lambda MM^{-1}| = |M(N - \lambda M^{-1})| = |M||N - \lambda M^{-1}| = |N - \lambda M^{-1}||M| = |(N - \lambda M_{-1})M| = |NM - \lambda M^{-1}M| = |NM - \lambda I|,$

etc. But if $M$ and $N$ are not inversible, then the proof above does not work. I found discussion in this link to be very relevant to my problem, but looks like they are using high tools that I am not familiar with. Is there any simple proof out there that I can digest well? As always, any gracious helping hand would be very much appreciated. Thank you. ~MA

 

[I like Serena]

How do I go about proving that for two matrices of same size, $M$ and $N$, that the characteristic polynomials of $MN$ and $NM$ are the same? If $M$ and $N$ are inversible, then the proof are very straightforward, for example, I can have

$|MN - \lambda I| = |MN - \lambda MM^{-1}| = |M(N - \lambda M^{-1})| = |M||N - \lambda M^{-1}| = |N - \lambda M^{-1}||M| = |(N - \lambda M_{-1})M| = |NM - \lambda M^{-1}M| = |NM - \lambda I|,$

etc. But if $M$ and $N$ are not inversible, then the proof above does not work. I found discussion in this link to be very relevant to my problem, but looks like they are using high tools that I am not familiar with. Is there any simple proof out there that I can digest well? As always, any gracious helping hand would be very much appreciated. Thank you. ~MA

Hey ~MA! ;)

Suppose $M$ is not invertible, then $(M+tI)$ is invertible for all sufficiently small values of $t$ that are non-zero (proof below).
So $(M+tI)N$ and $N(M+tI)$ have the same characteristic polynomial.
Now take the limit for $t\to 0$. (Thinking)

That leaves the question whether $M+tI$ is indeed invertible for all sufficiently small values of $t$ that are non-zero.
(That's what they mean when they say that all invertible nxn matrices are dense in the space of all nxn matrices.)
Consider that a matrix $M$ is non-invertible iff there is at least one non-zero vector $v$ such that $Mv=0$.
That means $0$ is an eigenvalue.
Since $(M+tI)v = Mv+tv=tv$, it follows that $t \ne 0$ is then an eigenvalue of $(M+tI)$.
Furthermore, as long as $|t|$ is smaller than the magnitude of the smallest non-zero eigenvalue, it follows in the same fashion that all eigenvalues are non-zero.
Therefore $(M+tI)$ is invertible for all sufficiently small values of $t$ that are non-zero. (Nerd)

 

[Opalg]

How do I go about proving that for two matrices of same size, $M$ and $N$, that the characteristic polynomials of $MN$ and $NM$ are the same? If $M$ and $N$ are inversible, then the proof are very straightforward, for example, I can have

$|MN - \lambda I| = |MN - \lambda MM^{-1}| = |M(N - \lambda M^{-1})| = |M||N - \lambda M^{-1}| = |N - \lambda M^{-1}||M| = |(N - \lambda M_{-1})M| = |NM - \lambda M^{-1}M| = |NM - \lambda I|,$

etc. But if $M$ and $N$ are not inversible, then the proof above does not work. I found discussion in this link to be very relevant to my problem, but looks like they are using high tools that I am not familiar with. Is there any simple proof out there that I can digest well? As always, any gracious helping hand would be very much appreciated. Thank you. ~MA

We want to show that if $MN - \lambda I$ is invertible then so is $NM - \lambda I$. There are two cases to look at.

Case 1: $\lambda = 0$. Here, the result follows from looking at the determinant. If $MN$ is invertible then $0 \ne |MN| = |M||N| = |N||M| =|NM|$, so that $NM$ is invertible.

Case 2: $\lambda\ne0$. Here, the result follows from a crafty piece of algebraic trickery. Suppose that $MN - \lambda I$ is invertible, and let $P = \lambda^{-1}(N(MN - \lambda I)^{-1}M- I).$ Then $$\begin{aligned} (NM - \lambda I)P &= \lambda^{-1}(NM - \lambda I) (N(MN - \lambda I)^{-1}M - I) \\ &= \lambda^{-1}NMN(MN - \lambda I)^{-1}M - N(MN - \lambda I)^{-1}M - \lambda^{-1}NM + I \\ &= \lambda^{-1}N(MN - \lambda I)(MN - \lambda I)^{-1}M - \lambda^{-1}NM + I \\ &= \lambda^{-1}NM - \lambda^{-1}NM + I \\ &= I.\end{aligned}$$ A similar calculation shows that $P(NM - \lambda I) = I$. It follows that $NM - \lambda I$ has an inverse, namely $P$.

 

[A.Magnus]

Hey ~MA! ;)

Suppose $M$ is not invertible, then $(M+tI)$ is invertible for all sufficiently small values of $t$ that are non-zero (proof below).
So $(M+tI)N$ and $N(M+tI)$ have the same characteristic polynomial.
Now take the limit for $t\to 0$. (Thinking)

That leaves the question whether $M+tI$ is indeed invertible for all sufficiently small values of $t$ that are non-zero.
(That's what they mean when they say that all invertible nxn matrices are dense in the space of all nxn matrices.)
Consider that a matrix $M$ is non-invertible iff there is at least one non-zero vector $v$ such that $Mv=0$.
That means $0$ is an eigenvalue.
Since $(M+tI)v = Mv+tv=tv$, it follows that $t \ne 0$ is then an eigenvalue of $(M+tI)$.
Furthermore, as long as $|t|$ is smaller than the magnitude of the smallest non-zero eigenvalue, it follows in the same fashion that all eigenvalues are non-zero.
Therefore $(M+tI)$ is invertible for all sufficiently small values of $t$ that are non-zero. (Nerd)

I am sorry, I am coming here late. But as always, thanks for your gracious helping hand! ~MA

 

[A.Magnus]

We want to show that if $MN - \lambda I$ is invertible then so is $NM - \lambda I$. There are two cases to look at.

Case 1: $\lambda = 0$. Here, the result follows from looking at the determinant. If $MN$ is invertible then $0 \ne |MN| = |M||N| = |N||M| =|NM|$, so that $NM$ is invertible.

Case 2: $\lambda\ne0$. Here, the result follows from a crafty piece of algebraic trickery. Suppose that $MN - \lambda I$ is invertible, and let $P = \lambda^{-1}(N(MN - \lambda I)^{-1}M- I).$ Then $$\begin{aligned} (NM - \lambda I)P &= \lambda^{-1}(NM - \lambda I) (N(MN - \lambda I)^{-1}M - I) \\ &= \lambda^{-1}NMN(MN - \lambda I)^{-1}M - N(MN - \lambda I)^{-1}M - \lambda^{-1}NM + I \\ &= \lambda^{-1}N(MN - \lambda I)(MN - \lambda I)^{-1}M - \lambda^{-1}NM + I \\ &= \lambda^{-1}NM - \lambda^{-1}NM + I \\ &= I.\end{aligned}$$ A similar calculation shows that $P(NM - \lambda I) = I$. It follows that $NM - \lambda I$ has an inverse, namely $P$.

Thanks to Opalg for your gracious helping hand. Now from this link I got another great hint, which is based on block (partition) matrices:

If $A$, $B$, $I$ and $0$ are all $n \times n$ matrices, and if
$$ M =
\begin{pmatrix}
I &0\\
-A &I\\
\end{pmatrix},
\text{ and }
N =
\begin{pmatrix}
\lambda I &B\\
\lambda A & \lambda I\\
\end{pmatrix},
$$
then
$$ MN =
\begin{pmatrix}
\lambda I &B\\
0 &\lambda I - AB\\
\end{pmatrix},
\text{ and }
NM =
\begin{pmatrix}
\lambda I - BA &B\\
0 & \lambda I\\
\end{pmatrix}.$$
Then based on $|M||N| = |MN| = |NM| = |N||M|$, we are able to prove our case. (But having found this link does not diminish my gratitude to all who have helped me graciously! Thanks again! ~MA)

While you are here, I have one question about multiplication of block matrices, if you don't mind: The determinant of |NM| is
$$\begin{vmatrix}
\lambda I - BA &B\\
0 & \lambda I\\
\end{vmatrix}
= (\lambda I - BA) \lambda I$$
and it is not $ \lambda I (\lambda I - BA)$, am I not correct? I am asking this question because matrix does not have commutative property, even though eventually the two come out to the same result. Please let me know and thank you again for your helping hand. ~MA

 

[I like Serena]

Thanks to Opalg for your gracious helping hand. Now from this link I got another great hint, which is based on block (partition) matrices:

If $A$, $B$, $I$ and $0$ are all $n \times n$ matrices, and if
$$ M =
\begin{pmatrix}
I &0\\
-A &I\\
\end{pmatrix},
\text{ and }
N =
\begin{pmatrix}
\lambda I &B\\
\lambda A & \lambda I\\
\end{pmatrix},
$$
then
$$ MN =
\begin{pmatrix}
\lambda I &B\\
0 &\lambda I - AB\\
\end{pmatrix},
\text{ and }
NM =
\begin{pmatrix}
\lambda I - BA &B\\
0 & \lambda I\\
\end{pmatrix}.$$
Then based on $|M||N| = |MN| = |NM| = |N||M|$, we are able to prove our case. (But having found this link does not diminish my gratitude to all who have helped me graciously! Thanks again! ~MA)

That gives us a third way to prove it! (Mmm)

While you are here, I have one question about multiplication of block matrices, if you don't mind: The determinant of |NM| is
$$\begin{vmatrix}
\lambda I - BA &B\\
0 & \lambda I\\
\end{vmatrix}
= (\lambda I - BA) \lambda I$$
and it is not $ \lambda I (\lambda I - BA)$, am I not correct? I am asking this question because matrix does not have commutative property, even though eventually the two come out to the same result. Please let me know and thank you again for your helping hand. ~MA

Shouldn't that be:
$$\begin{vmatrix}
\lambda I - BA &B\\
0 & \lambda I\\
\end{vmatrix}
= |\lambda I - BA| \cdot |\lambda I| = |\lambda I| \cdot |\lambda I - BA|$$
(Wondering)

 

[A.Magnus]

That gives us a third way to prove it! (Mmm)

Shouldn't that be:
$$\begin{vmatrix}
\lambda I - BA &B\\
0 & \lambda I\\
\end{vmatrix}
= |\lambda I - BA| \cdot |\lambda I| = |\lambda I| \cdot |\lambda I - BA|$$
(Wondering)

You maybe right because determinant is scalar and it enjoys commutative property, but I would like to know multiplication of block matrices in general case, for example,

$$\begin{pmatrix}
A &B\\
\end{pmatrix}
\begin{pmatrix}
C\\
D\\
\end{pmatrix}
=
\begin{pmatrix}
AC + BD
\end{pmatrix}
\neq
\begin{pmatrix}
CA + DB\\
\end{pmatrix},$$

am I not correct? I am referring to the non-commutative property of matrix. Any idea? Thanks to all for your attention and gracious help! ~MA

 

[I like Serena]

That is correct.