Eigen Value

Question

An orthogonal matrix A has eigen values 1, 2 and 4. What is the trace of the matrix 

Comments

For a matrix 'A' to be Orthogonal, it should satisfy

AA^T  = I (Identity Matrix)

we can simplify it as :

==> A^-1 AA^T = A^-1 I

==> I A^T = A^-1 I

==> A^T = A^-1

We know the Eigen Values of A <1, 2, 4>, therfore Eigen values of A^-1 are <1, 1/2, 1/4>

Trace = 1 + 1/2 + 1/4

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yeah you're correct
https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors#Additional_properties_of_eigenvalues

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I think the question is not correct..

I added my answer

Answer 1

Given question is wrong.

if A is an orthogonal matrix then $det A=\pm 1$

Proof: 

$\because AA^t=I$

$\Rightarrow |A A^t|=|I|$

$\Rightarrow |A| |A^t|=1$

$\Rightarrow|A|^2=1$

$\Rightarrow  |A|=\pm 1$

 

But in the given question-

$det \; A= 1*2*4 (\text{product of eigen values}) =8 \neq \pm 1$

So, the given matrix is not an orthogonal matrix.

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Matrix $A$ and $A^t$ have the same eigen values.

In case of orthogonal matrix, $AA^{t} = I$ 

Pre-multiplying by $A^{-1}$ both sides,

we get, $A^{t}=A^{-1}$

$\implies $eigen values of $A$ = eigen values of $A^t$= eigen values of $A^{-1}$

So the Trace of these matrices should also be equal in case of orthogonal matrices.

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Proof $1$ :- Matrices $A$ and $A^{t}$ have the same eigen values.

If $\lambda$ is an eigen value of a square matrix $A$

then $|A-\lambda I| = 0$

$\Rightarrow |(A-\lambda I)^{t}|=0$

$\Rightarrow |(A^t-\lambda I)|=0$

So, $\lambda$ should also be an eigen value of a square matrix $A^{t}.$

Proof $2$ :- If $\lambda$ is an eigen value of a square matrix $A$ then  $1/ \lambda$ is an eigen value of a square matrix $A^{-1}$

$AX=\lambda X$

Pre-multiplying by $A^{-1}$ both sides,

$\Rightarrow X=A^{-1}\lambda X$

$\Rightarrow X=\lambda A^{-1} X$

$\Rightarrow \frac{1}{\lambda}X= A^{-1} X$

So, $1/ \lambda$ should be an eigen value of a square matrix $A^{-1}$

Answer 2

For a matrix 'A' to be Orthogonal, it should satisfy

AA^T  = I (Identity Matrix)

we can simplify it as :

==> A^-1 AA^T = A^-1 I

==> I A^T = A^-1 I

==> A^T = A^-1

We know the Eigen Values of A <1, 2, 4>, therefore Eigen values of A^-1 are <1, 1/2, 1/4>

Trace = 1 + 1/2 + 1/4

Comments

matrices $A$ and $A^T$   have same eigen values.

is this statement wrong??

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No.

Answer 3

Trace of matrix is equal to the sum of eigen values i.e. 1+2+4=7

Answer 4

wrong question