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A matrix satisfies A^2 - 4A + 5I = 0, then n is even.

by Hydroxide
Tags: determinants
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Hydroxide
#1
Apr21-07, 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(AT x A) = 0

3. The attempt at a solution

1) (A-2I)^2 +I=0

Not sure what to do after this though


Thanks in advance
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AKG
#2
Apr21-07, 10:24 AM
Sci Advisor
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P: 2,586
Quote Quote by Hydroxide View Post
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.
An n x n matrix with real entries? If it can have complex entries, this isn't true. Anyways, what do you know about minimal polynomials and characteristic polynomials?
2) Let A be an m x n matrix where m<n. Show that det(A^T x A) = 0
What do you know about the relationships between rank, invertibility, and determinants.
1) (A-2I)^2 +I=0
Okay, that's not bad. So (A-2I)2 = -I. Compute the determinant of both sides.
Hydroxide
#3
Apr21-07, 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.

Quote Quote by AKG View Post
What do you know about the relationships between rank, invertibility, and determinants.
Could you explain further please?

AKG
#4
Apr21-07, 11:30 PM
Sci Advisor
HW Helper
P: 2,586
A matrix satisfies A^2 - 4A + 5I = 0, then n is even.

What are the dimensions of the matrix ATA? What can you say about the rank of ATA?
Hydroxide
#5
Apr22-07, 12:43 AM
P: 16
Quote Quote by AKG View Post
What are the dimensions of the matrix ATA? What can you say about the rank of ATA?
ATA is n x n
We haven't covered ranks yet

I know that ATA 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.
Glass
#6
Apr22-07, 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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