Show that eigenvalue of A + eigvalueof B ≠ eigvalue of A+B?

[Oliviacarone]

Homework Statement

Let A and B be nxn matrices with Eigen values λ and μ, respectively.
a) Give an example to show that λ+μ doesn't have to be an Eigen value of A+B
b) Give an example to show that λμ doesn't have to be an Eigen value of AB

Homework Equations

det(λI - A)=0

The Attempt at a Solution

I tried a bunch of matrices with negatives, some with Eigen value of 0, but none of them seem to work. For example, I tried [20 02] and [60 06], or [01 01] and [20 02], which both didn't work. Please help!

 

[fresh_42]

Homework Statement

Let A and B be nxn matrices with Eigen values λ and μ, respectively.
a) Give an example to show that λ+μ doesn't have to be an Eigen value of A+B
b) Give an example to show that λμ doesn't have to be an Eigen value of AB

Homework Equations

det(λI - A)=0

The Attempt at a Solution

I tried a bunch of matrices with negatives, some with Eigen value of 0, but none of them seem to work. For example, I tried [20 02] and [60 06], or [01 01] and [20 02], which both didn't work. Please help!

Always start with what you have. Simply list it:
$A.v=\lambda \cdot v$ and $B.w=\mu \cdot w$.
This already gives you a hint, if you try to write $(A+B).v = ?\; , \;(A+B).w =?\; , \;(AB).w = \mu A.w= ?$

 

[Oliviacarone]

Always start with what you have. Simply list it:
$A.v=\lambda \cdot v$ and $B.w=\mu \cdot w$.
This already gives you a hint, if you try to write $(A+B).v = ?\; , \;(A+B).w =?\; , \;(AB).w = \mu A.w= ?$

I'm not sure what you're getting at,
I did (A+B)v = Av +Bv = λv + Bv
(A+B)w = Aw + Bw = Aw + μw

Not sure what this gives me though :/

 

[fresh_42]

I'm not sure what you're getting at,
I did (A+B)v = Av +Bv = λv + Bv
(A+B)w = Aw + Bw = Aw + μw

Not sure what this gives me though :/

The hint was $v \neq w$. Why do you assume them to be the same?

 

[Oliviacarone]

The hint was $v \neq w$. Why do you assume them to be the same?

What did I write that made me assume they're the same? I know they're different eigenvectors, I want to know why they can be different eigenvalues

 

[fresh_42]

What did I write that made me assume they're the same? I know they're different eigenvectors, ...

Wrong. You used only $v$, resp. $w$ for both matrices (in post #3), which means you implicitly used $v=w$.

... I want to know why they can be different eigenvalues

By using the same eigenvector, you won't find a counterexample.

 

[Oliviacarone]

Wrong. You used only $v$, resp. $w$ for both matrices (in post #3), which means you implicitly used $v=w$.

By using the same eigenvector, you won't find a counterexample.

I'm sorry I honestly don't understand what you mean by saying I only used v.
When I wrote (A+B)w= Aw+Bw, aren't I using w??

 

[fresh_42]

I'm sorry I honestly don't understand what you mean by saying I only used v.
When I wrote (A+B)w= Aw+Bw, aren't I using w??

I gave $A.v=\lambda v$ and you used $A.w=\lambda w$. What you actually have is $(A+B).v= \lambda v + B.v$ and you know nothing about $B.v$, or you have $(A+B).w=A.w + \mu w$ and you know nothing about $A.w$. So the trick is to look for an example like $A.\begin{bmatrix}1\\0\end{bmatrix}=2\cdot \begin{bmatrix}1\\0\end{bmatrix}$ and $B.\begin{bmatrix}0\\1\end{bmatrix}= 3\cdot \begin{bmatrix}0\\1\end{bmatrix}$ with (almost) whatever you like for $A.\begin{bmatrix}0\\1\end{bmatrix}$ and $B.\begin{bmatrix}1\\0\end{bmatrix}$.

 

[Oliviacarone]

I gave $A.v=\lambda v$ and you used $A.w=\lambda w$. What you actually have is $(A+B).v= \lambda v + B.v$ and you know nothing about $B.v$, or you have $(A+B).w=A.w + \mu w$ and you know nothing about $A.w$. So the trick is to look for an example like $A.\begin{bmatrix}1\\0\end{bmatrix}=2\cdot \begin{bmatrix}1\\0\end{bmatrix}$ and $B.\begin{bmatrix}0\\1\end{bmatrix}= 3\cdot \begin{bmatrix}0\\1\end{bmatrix}$ with (almost) whatever you like for $A.\begin{bmatrix}0\\1\end{bmatrix}$ and $B.\begin{bmatrix}1\\0\end{bmatrix}$.

Okay... So I found A = [2 1 0 1] and B = [1 0 1 3] which fits your equations, but these both have 2 Eigen values each, which isn't what I'm looking for... really not sure what you're trying to say

 

[fresh_42]

Let A and B be nxn matrices with Eigen values λ and μ, respectively.

There is absolutely no word about the corresponding eigenvectors. If you assume both matrices to have the same eigenvector $v$, then you will necessarily get $(A+B).v=(\lambda +\mu)\cdot v$ and $(AB)=\lambda \mu \cdot v$, which is not what's requested.
Therefore you will have to select two different eigenvectors for both eigenvalues, plus make sure, that the other one isn't also an eigenvector of the other matrix.

So the question is not why I've chosen two different eigenvectors, but why you insist to do not?

 

[Oliviacarone]

Okay, so I get that, but they could still have (λ+μ) = Eigen value of (A+B) no ?

 

[fresh_42]

Okay, so I get that, but they could still have (λ+μ) = Eigen value of (A+B) no ?

This could happen, but you're requested to find an example where it doesn't (which happens more often).

 

[Oliviacarone]

This could happen, but you're requested to find an example where it doesn't (which happens more often).

So my answer is the A and B i gave + that the Eigen vectors are different so it doesn't matter if the (A+B) Eigen value is the same as the Eigen value A + Eigen value B?

 

[Oliviacarone]

So my answer is the A and B i gave + that the Eigen vectors are different so it doesn't matter if the (A+B) Eigen value is the same as the Eigen value A + Eigen value B?

because if I just answer it with that A and B, the Eigen values are 2 and 1 for A, and 1 and 3 for B. For AB, the Eigen values are 3 and 4 so Eigen value A + Eigen value B can = Eigen value A+B

 

[fresh_42]

So my answer is the A and B i gave + that the Eigen vectors are different so it doesn't matter if the (A+B) Eigen value is the same as the Eigen value A + Eigen value B?

Yes. If you take $A.(1,0)^\tau= 2\cdot (1,0)^\tau$ and $B.(0,1)^\tau=3 \cdot (0,1)^\tau$ then $\lambda = 2$ and $\mu =3$ and neither $2+3=5$ nor $2\cdot 3=6$ are eigenvalues of $A+B\; , \;AB\,,$ resp., or are they? I haven't checked.

 

[PeroK]

So my answer is the A and B i gave + that the Eigen vectors are different so it doesn't matter if the (A+B) Eigen value is the same as the Eigen value A + Eigen value B?

Here's another way to look at it. Take 2x2 matrices. Suppose A has eigenvalues 1, 2 and B has eigenvalues 10 and 20, then A + B must have eigenvalues 11, 12, 21, 22? Or not?

 

[fresh_42]

because if I just answer it with that A and B, the Eigen values are 2 and 1 for A, and 1 and 3 for B. For AB, the Eigen values are 3 and 4 so Eigen value A + Eigen value B can = Eigen value A+B

You have to consider all combinations: $\{3,5,2,4\}$, resp. $\{2,6,1,3\}$. But it's not difficult to compute the eigenvalues of $A+B$ and $AB$ in your example.

 

[Oliviacarone]

Yes. If you take $A.(1,0)^\tau= 2\cdot (1,0)^\tau$ and $B.(0,1)^\tau=3 \cdot (0,1)^\tau$ then $\lambda = 2$ and $\mu =3$ and neither $2+3=5$ nor $2\cdot 3=6$ are eigenvalues of $A+B\; , \;AB\,,$ resp., or are they? I haven't checked.

No, you're right (at least for A+B, I haven't checked for AB e

You have to consider all combinations: $\{3,5,2,4\}$, resp. $\{2,6,1,3\}$. But it's not difficult to compute the eigenvalues of $A+B$ and $AB$ in your example.

Okay yeah so then those don't work? A= 2 and 1, B= 3 and 4. A+B = 1 and 3, and 2+1 =3

 

[Oliviacarone]

Here's another way to look at it. Take 2x2 matrices. Suppose A has eigenvalues 1, 2 and B has eigenvalues 10 and 20, then A + B must have eigenvalues 11, 12, 21, 22? Or not?

hmmm no, only 11 and 22

 

[PeroK]

hmmm no, only 11 and 22

Hmmm that's an interesting answer!

 

[Oliviacarone]

Hmmm that's an interesting answer!

Okay so I can just answer this question with
A = [1 0 0 2] and B = [10 0 0 20] and write out all the Eigen values and that there's no Eigen value of 12 or 21 for A+B?
To me, the question was saying that A and B only had one Eigen value each that's why I was confused

 

[fresh_42]

No, you're right (at least for A+B, I haven't checked for AB e

Okay yeah so then those don't work? A= 2 and 1, B= 3 and 4. A+B = 1 and 3, and 2+1 =3

Your matrices $A=\begin{bmatrix}2&1\\0&1\end{bmatrix}\; , \;B=\begin{bmatrix}1&0\\1&3\end{bmatrix}$ work perfectly for the exercise.
It's the diagonal matrices which would cause problems. (Although the diagonal matrices are those who are usually fine, because of their behavior with eigenvalues and eigenvectors. Just not as a counterexample here.)

 

[Oliviacarone]

Your matrices $A=\begin{bmatrix}2&1\\0&1\end{bmatrix}\; , \;B=\begin{bmatrix}1&0\\1&3\end{bmatrix}$ work perfectly for the exercise.
It's the diagonal matrices which would cause problems. (Although the diagonal matrices are those who are usually fine, because of their behavior with eigenvalues and eigenvectors. Just not as a counterexample here.)

But these have Eigen values of 2, 1 for A and 3,1 for B. A+B has Eigen values of 3 and 4, which can be found by adding Eigen values of A and B. BUT A and B also gives other Eigen values (2+3 = 5, for example) ... is this why it works??

 

[fresh_42]

But these have Eigen values of 2, 1 for A and 3,1 for B.

correct.

A+B has Eigen values of 3 and 4

not correct

which can be found by adding Eigen values of A and B. BUT A and B also gives other Eigen values (2+3 = 5, for example) ... is this why it works??

The false statements are:
If $A$ has an eigenvalue $\lambda$, that is $A.v=\lambda \cdot v$ for an eigenvector $v\neq 0$, and $B$ has an eigenvalue $\mu$, that is $B.w=\mu \cdot w$ for an eigenvector $w\neq 0$, then
a) [False:] $A+B$ has an eigenvalue $\lambda + \mu$
b) [False:] $A\cdot B$ has an eigenvalue $\lambda \cdot \mu$

Find an example for $A$ and $B$ which shows, that these are wrong. And your example does the job - at least the version I wrote in post #22. Your one-line notation wasn't quite clear. Just compute the eigenvalues of $A+B$ and $A\cdot B$ correctly, and you will see.

You should also calculate an example with diagonal matrices, in order to see, why they are no counterexample.

 

[Oliviacarone]

correct.not correct
The false statements are:
If $A$ has an eigenvalue $\lambda$, that is $A.v=\lambda \cdot v$ for an eigenvector $v\neq 0$, and $B$ has an eigenvalue $\mu$, that is $B.w=\mu \cdot w$ for an eigenvector $w\neq 0$, then
a) [False:] $A+B$ has an eigenvalue $\lambda + \mu$
b) [False:] $A\cdot B$ has an eigenvalue $\lambda \cdot \mu$

Find an example for $A$ and $B$ which shows, that these are wrong. And your example does the job - at least the version I wrote in post #22. Your one-line notation wasn't quite clear. Just compute the eigenvalues of $A+B$ and $A\cdot B$ correctly, and you will see.

You should also calculate an example with diagonal matrices, in order to see, why they are no counterexample.

I totally understand now. Thank you so much. Now, if the Eigen vectors HAD been the same, would I just prove this by saying (A+B)v = (λ+μ)v ?

 

[fresh_42]

Now, if the Eigen vectors HAD been the same, would I just prove this by saying (A+B)v = (λ+μ)v ?

You must be more precise here. For arbitrary matrices or for just a certain pair $A,B$? But yes, if $A$ and $B$ have an eigenvector $v$ in common to eventually (but not necessarily) different eigenvalues $\lambda ,\mu$, then you can write this equation to show that the sum has at least one (possibly more than one, possibly not) eigenvector $v$ to the added eigenvalues. E.g. diagonal matrices have this property. Also $A,B^\tau$ would have one eigenvector in common (but not both). The fact that one is upper and the other lower triangular saved your day.

 

[Oliviacarone]

correct.not correct
The false statements are:
If $A$ has an eigenvalue $\lambda$, that is $A.v=\lambda \cdot v$ for an eigenvector $v\neq 0$, and $B$ has an eigenvalue $\mu$, that is $B.w=\mu \cdot w$ for an eigenvector $w\neq 0$, then
a) [False:] $A+B$ has an eigenvalue $\lambda + \mu$
b) [False:] $A\cdot B$ has an eigenvalue $\lambda \cdot \mu$

Find an example for $A$ and $B$ which shows, that these are wrong. And your example does the job - at least the version I wrote in post #22. Your one-line notation wasn't quite clear. Just compute the eigenvalues of $A+B$ and $A\cdot B$ correctly, and you will see.

You should also calculate an example with diagonal matrices, in order to see, why they are no counterexample.

I totally understand now. Thank you so much. Now, if the Eigen vectors HAD been the same, how would I prove this

Hmmm that's an interesting answer!

You must be more precise here. For arbitrary matrices or for just a certain pair $A,B$? But yes, if $A$ and $B$ have an eigenvector $v$ in common to eventually (but not necessarily) different eigenvalues $\lambda ,\mu$, then you can write this equation to show that the sum has at least one (possibly more than one, possibly not) eigenvector $v$ to the added eigenvalues. E.g. diagonal matrices have this property. Also $A,B^\tau$ would have one eigenvector in common (but not both). The fact that one is upper and the other lower triangular saved your day.

It says: Prove that if λ and μ correspond to the same Eigen vector x then λ+μ is an Eigen value of A+B and λμ is an eigenvalue of AB.
So that equation would prove this?

 

[fresh_42]

It says: Prove that if λ and μ correspond to the same Eigen vector x then λ+μ is an Eigen value of A+B and λμ is an eigenvalue of AB.
So that equation would prove this?

Yes. And likewise for $AB.x=(AB)(x)=A(B(x))= A(\mu \cdot x)=\mu \cdot A(x) = \ldots$ if you write it in great detail.

 

[Oliviacarone]

Yes. And likewise for $AB.x=(AB)(x)=A(B(x))= A(\mu \cdot x)=\mu \cdot A(x) = \ldots$ if you write it in great detail.

Thank you!