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

In summary, the conversation discusses two problems involving matrices. The first problem states that for an n x n matrix A satisfying the equation A^2 - 4A + 5I = 0, n must be even. The second problem states that for an m x n matrix A with m < n, the determinant of A^T x A must be equal to 0. The conversation also mentions using minimal polynomials and characteristic polynomials in the first problem, as well as exploring the relationships between rank, invertibility, and determinants in the second problem. The conversation ends with a hint to consider matrix multiplication as a linear combination of columns.
  • #1
Hydroxide
16
0

Homework Statement



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

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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  • #2
Hydroxide said:

Homework Statement



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.
 
  • #3
Cheers I've got the first question now. Was easier than i thought.

I still can't do 2) though.

AKG said:
What do you know about the relationships between rank, invertibility, and determinants.

Could you explain further please?
 
  • #4
What are the dimensions of the matrix ATA? What can you say about the rank of ATA?
 
  • #5
AKG said:
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.
 
  • #6
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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1. What is a matrix?

A matrix is a rectangular array of numbers or symbols arranged in rows and columns. It is often used to represent linear transformations or systems of linear equations.

2. What does it mean for a matrix to satisfy A^2 - 4A + 5I = 0?

When a matrix satisfies this equation, it means that when the matrix is squared and then multiplied by 4 and 5 times the identity matrix, the resulting matrix will be equal to the zero matrix. In other words, the matrix has a special relationship with its own square and the identity matrix.

3. How do you solve for n when a matrix satisfies A^2 - 4A + 5I = 0?

To solve for n, we can use the fact that A^2 represents the matrix multiplied by itself. So, we can rewrite the equation as A(A - 4I) = -5I. Since the determinant of A is equal to the product of its eigenvalues and the determinant of a scalar multiple of a matrix is equal to the scalar multiple of the determinant, we can say that det(A) * det(A - 4I) = -5. Since the determinant of the identity matrix is 1, we know that det(A) * (det(A) - 4) = -5. We can then solve for n by finding the values of det(A) that satisfy this equation.

4. What does it mean for n to be even in this context?

In this context, n represents the size (or order) of the matrix. When n is even, it means that the matrix has an even number of rows and columns. This is important because it affects the solution to the equation A^2 - 4A + 5I = 0.

5. Can a matrix satisfy A^2 - 4A + 5I = 0 if n is odd?

No, a matrix cannot satisfy this equation if n is odd. This is because when n is odd, the matrix will have an odd number of rows and columns, and the equation only holds true when n is even. This is because when a matrix is squared, its size is doubled, and if the original matrix has an odd number of rows and columns, it will never be possible to subtract 4A without changing the size of the matrix to an even number.

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