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Eigenvalues and eigenvectors

Yankel

Active member
Jan 27, 2012
398
One more question please...

which one of these statements is NOT true (only one can be false):

a. a square matrix nXn with n different eigenvalues can become diagonal.

b. A matrix that can be diagonal is irreversible.

c. Eigenvectors that correspond to different eigenvalues are linearly independent.

d. There are square matrices with no real eigenvalue.

I think that b is correct....

thanks
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
One more question please...

which one of these statements is NOT true (only one can be false):

a. a square matrix nXn with n different eigenvalues can become diagonal.

b. A matrix that can be diagonal is irreversible.

c. Eigenvectors that correspond to different eigenvalues are linearly independent.

d. There are square matrices with no real eigenvalue.

I think that b is correct....

thanks
What does "irreversible" mean in this context? Do you mean invertible?
 

Yankel

Active member
Jan 27, 2012
398
yes, sorry, I mean not invertible, meaning, has no inverse.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
consider this: the matrix I is diagonal, yet it is invertible, so....
 

Yankel

Active member
Jan 27, 2012
398
I have eliminated most answers out, so I am left with 2....

the first is the invertible thing, and the second is that there are no squared matrices with no real eignvalue.

According to definition: D=P^-1 * A * P

So does A need to be invertible ? Why ?

Is it possible that a characteristic polynomial will have no real roots for a squared matrix ? I think so...
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
I have eliminated most answers out, so I am left with 2....

the first is the invertible thing, and the second is that there are no squared matrices with no real eignvalue.

According to definition: D=P^-1 * A * P

So does A need to be invertible ? Why ?

Is it possible that a characteristic polynomial will have no real roots for a squared matrix ? I think so...
How about
$$\begin{bmatrix}0 &-1\\ 1 &0\end{bmatrix}?$$
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
How about
$$\begin{bmatrix}0 &-1\\ 1 &0\end{bmatrix}?$$
And just to be clear about the second option, you mean to write that

b. A matrix that can be diagonalized is non-invertible (or singular).

Is that the correct b. option?
 

Yankel

Active member
Jan 27, 2012
398
I don't know which one is correct, but I am quite convinced now it's b, your example with the 2X2 matrix was good, I think it's done, thanks !!
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
I don't know which one is correct, but I am quite convinced now it's b, your example with the 2X2 matrix was good, I think it's done, thanks !!
You're welcome!
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
I don't know which one is correct, but I am quite convinced now it's b, your example with the 2X2 matrix was good, I think it's done, thanks !!
as i pointed out before, I is a diagonal matrix (thus it is "diagonalizable" you don't even have to do anything!) that is invertible (I is its own inverse), so b. must be false.