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3 Show that two square matrices A and B of order n are similar if and only if their eigenvalues coincide and for each eigenvalue A and each i < n have rank(A AE)...

Question

3 Show that two square matrices A and B of order n are similar if and only if their eigenvalues coincide and for each eigenvalue A and each i < n have rank(A AE)i = rank(B AE)' .

3 Show that two square matrices A and B of order n are similar if and only if their eigenvalues coincide and for each eigenvalue A and each i < n have rank(A AE)i = rank(B AE)' .



Answers

(a) Prove that if $A$ is a square matrix, then $A$ and $A^{T}$ have the same eigenvalues. [Hint: Look at the characteristic equation $\operatorname{det}(\lambda I-A)=0 .]$ (b) Show that $A$ and $A^{T}$ need not have the same eigenspaces. [Hint: Use the result in Exercise 30 to find a $2 \times 2$ matrix for which $\left.A \text { and } A^{T} \text { have different eigenspaces. }\right]$

I miss probably already know that both A and B s symmetric metrics with polity Begum values that news for any X In our in we have x transpose times say comes eggs is positive and the x transpose times b times x is positive. This is this court this metric smart applications are linear so we have X transpose times a prosperity Tom sakes. He's supposed tiff that is on the scene The quadratic form defined by this way cure eggs equals to exchange suppose tubs a plus B time. Sex is hold here definite, which tells us the new metrics a process be has positive Eigen values.

Things exercise. We will assume that fee is a Nuyorican factor for a with corresponding Eigen value off Lambda That is that eight times three equals Lambda V on. We will also assume that he is a similar similar to a on that means that we can be written as its inverse times eight s for some metrics is on. We have to prove what it's written in green. So we will start by writing be times is immersed in this fee on by replacing B die as embarrassed times A times s this way. Um on because the product of matrices is associative We can rewrite this expression as s teams as in verse and we can do that product first, which is the identity matrix. So this gives us as immersed in a team's fee on Dhere. We use our first hypothesis that if we multiply eight msv, we obtained Lambda Fee. So since Lambda is the scaler, we can put it at the front. So this is the same as Lambda This embarrass fee on. Now, if you look carefully, we obtained what we wanted to prove that the times as inverse times fee equals Lambda teams is inverse times way. Which means that ISS inverse times fee is an icon. Victor off be on this finishes the exercise.


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