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A matrix satisfies A^2  4A + 5I = 0, then n is even.by Hydroxide
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#1
Apr2107, 02:39 AM

P: 16

1. The problem statement, all variables and given/known data
1) Let A be an n x n matrix with A^2 4A +5I = 0. Show that n must be even. 2) Let A be an m x n matrix where m<n. Show that det(A^{T} x A) = 0 3. The attempt at a solution 1) (A2I)^2 +I=0 Not sure what to do after this though Thanks in advance 


#2
Apr2107, 10:24 AM

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#3
Apr2107, 09:59 PM

P: 16

Cheers I've got the first question now. Was easier than i thought.
I still can't do 2) though. 


#4
Apr2107, 11:30 PM

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A matrix satisfies A^2  4A + 5I = 0, then n is even.
What are the dimensions of the matrix A^{T}A? What can you say about the rank of A^{T}A?



#5
Apr2207, 12:43 AM

P: 16

We haven't covered ranks yet I know that A^{T}A can reduced so that it has one row of zero's hence det=0. But I don't know how to show it in general. 


#6
Apr2207, 03:38 AM

P: 24

It may help to think of matrix multiplication with a vector as a linear combination of the columns of the matrix
i.e. For [tex]A\vec{c} = \vec{b}\\[/tex] b is a linear combination of the columns of A And hence a matrix multiplication with a vector will produce a matrix whose columns are a linear combination of the columns of the first matrix. i.e. For [tex]AB = C\\[/tex] C's columns are linear combinations of the columns of A Sorry if the Latex is less than desirable, as you can see, I'm new here. 


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