Solving Matrices Questions: Trace of a Matrix [tr(A)]

[Tanishq Nandan]

Homework Statement

Homework Equations

A.A^-1=Identity matrix
Trace of a matrix [tr(A)]is the sum of elements on it's main diagonal.

The Attempt at a Solution

In the given equation,post-multiplying A^-1 (A inverse) on both sides gives A^4=16.
Since the array contains only one element (say x),this can only mean that x^4=16,
which means that x is either 2 or -2.So,the trace of the matrix can also correspondingly be 2 or -2
Sure enough,the answer given is 2,but how do we eliminate -2?
There's nothing in the question which says the matrix elements can't be negative.
That would mean I'm most probably missing something basic.
Help appreciated.

 

[scottdave]

Can you get A⁵ = 16A by using -2 ?

 

[Tanishq Nandan]

Umm...yeah..why not??
-32=-32

 

[StoneTemplePython]

The problem is kind of hard to read / interpret to be honest.

Why do you think $\mathbf A^{-1}$ exists?

 

[Mark44]

Homework Statement

View attachment 210415

Homework Equations

A.A^-1=Identity matrix
Trace of a matrix [tr(A)]is the sum of elements on it's main diagonal.

The Attempt at a Solution

In the given equation,post-multiplying A^-1 (A inverse) on both sides gives A^4=16.

You're not given that matrix A is invertible, so $A^{-1}$ is not guaranteed to exist.

Since the array contains only one element (say x),this can only mean that x^4=16,

No, I don't think this is true. As I read the problem, A is an n x n matrix, each of whose elements is the product of two real numbers.

The problem wording seems a bit odd to me.

which means that x is either 2 or -2.So,the trace of the matrix can also correspondingly be 2 or -2
Sure enough,the asnwer given is 2,but how do we eliminate -2?
There's nothing in the question which says the matrix elements can't be negative.
That would mean I'm most probably missing something basic.
Help appreciated.

 

[Tanishq Nandan]

Yeah,valid point..guarantee for A^-1 to exist.

s I read the problem, A is an n x n matrix, each of whose elements is the product of two real numbers.

Yep,with n=1 to make it simple

The problem wording seems a bit odd to me.

I know..the first part went entirely above me.
I just did whatever I could from the given equation.
That too,apparently turned out to be wrong.

 

[Mark44]

Yep,with n=1 to make it simple

When you're working with matrices, it's better to work with at least 2 x 2 matrices.

You are given that $A^5 = 16A$, which is equivalent to $A(A^4 - 16I) = 0$. I would start be factoring the left side completely, but not assuming any particular size for the matrices.

 

[Tanishq Nandan]

not assuming any particular size for the matrices.

But the question has already said that the matrix consists of a single element,a(ij)

 

[Tanishq Nandan]

A is a singleton matrix.I thought that's what the first part meant.

 

[Mark44]

But the question has already said that the matrix consists of a single element,a(ij)

No.

The notation $[a_{ij}]$, with $1 \le i \le n$ and $1 \le j \le n$ means that the matrix A is n X n. At least, that's the usual meaning. I don't get the connection with u and v, though.

 

[Tanishq Nandan]

Oo..alright

I don't get the connection with u and v, though.

Me neither

 

[StoneTemplePython]

I think what they're saying is that $\mathbf A$ is a rank one matrix in reals, i.e.

$\mathbf A = \mathbf {uv}^T$

hence

$trace\big(\mathbf A\big) = trace\big(\mathbf {uv}^T\big) = trace\big(\mathbf {v}^T\mathbf u \big)=\lambda$

If you're not familiar with diagonalization, you can attack this just with extensive use of cyclic property of trace. However, the problem seems underspecified.

 

[Tanishq Nandan]

I don't think I got that..

 

[SammyS]

I think what they're saying is that $\mathbf A$ is ...

$\mathbf A = \mathbf {uv}^T$

Yes. That's how I read this problem. u and v are column vectors.

 

[StoneTemplePython]

Yes. That's how I read this problem.

The issue is it's a terrible problem.

suppose

$\mathbf u = \mathbf 0$
Then

$\mathbf A = \mathbf {uv}^T= \mathbf{0}$

i.e. $\mathbf A$ is the zero matrix which is in fact rank zero. This allows $\lambda = 0$ as a solution. To get the 'official' solution of only $\lambda = 2$, it needs to say that all entries in $\mathbf u$ and $\mathbf v$ are positive or some other condition that is considerably more restrictive than the original question.

 

[StoneTemplePython]

I don't think I got that..

Are you familiar with the cyclic property of the trace? If not, I'd drill that as it's quite important in many different situations and can give you insights to this problem.

As is, the problem you posted with the solution of just $\lambda =2$, is broken.

 

[Tanishq Nandan]

Ok,I'll look into it.

 

[I like Serena]

Homework Statement

View attachment 210415

Homework Equations

A.A^-1=Identity matrix
Trace of a matrix [tr(A)]is the sum of elements on it's main diagonal.

The Attempt at a Solution

In the given equation,post-multiplying A^-1 (A inverse) on both sides gives A^4=16.
Since the array contains only one element (say x),this can only mean that x^4=16,
which means that x is either 2 or -2.So,the trace of the matrix can also correspondingly be 2 or -2
Sure enough,the answer given is 2,but how do we eliminate -2?
There's nothing in the question which says the matrix elements can't be negative.
That would mean I'm most probably missing something basic.
Help appreciated.

You haven't missed anything - you're right on target.
With u=(1) and v=(2), we get A=(2), which satisfies all conditions, and which has trace 2.
With u=(1) and v=(-2), we get A=(-2), which also satisfies all conditions, and which has trace -2.
Just for fun, with u=(0) and v=(0), we get A=(0), which also satisfies all conditions, and which has trace 0.
In other words, with the given information we cannot tell what the trace is, and we have a counter example for the given answer.

Going a little beyond, the given equation implies that we can only have eigenvalues -2, 0, and 2 (and pairs of -2i and +2i that cancel), all with unknown multiplicities.
The trace is the sum of the eigenvalues, so I believe the only thing we can say is that the trace is an even integer.

Just checking, is there any additionally information given before the whole set of problem statements?

 

[Tanishq Nandan]

Nope