Anonymous

Just verifying.?

From the question If a is the eigenvalue of A, then a is also the eigenvalue of AT where T means transpose.

Proof (I've tried answering it not sure if I am right)

If Ax=ax by definition

then Ax-ax=0

=> (A-aI)x=0

=> (A-aI)Tx=0T

=> (AT-aIT)x=0 (since a is just a scalar from the properties of the transpose (aA)T=aAT and IT=I

Thus, ATx=ax. completes the proof!

Relevance

seems correct to me

Although I don't like to use the notation T for transpose,

I will just stick to the notation (I would use "^t" as in A^t).

Let us look at the following arguments you had:

(A - aI)x = 0 => (A - aI)Tx = 0T.

To go from LHS to RHS, I think that you transpose the

both sides, but the transpose of (A - aI)x is not (A - aI)Tx.

Why? In general, (AB)T is not ATBT. In general, (AB)T

= BTAT.

Now do you see how you can fix the proof?

Let me add few more lines since some people seem confused.

When the questioner deduced (A - aI)Tx = 0T (3rd line from the

deductions) from the second line (A - aI)x = 0, the questioner

was trying to transpose the left hand side and the right hand

side. However, the transpose of (A - aI)x is not (A - aI)Tx as

the questioner wrongfully concluded. Let M = A - aI and N = x

be two matrices, then as I explained above (MN)T = NTMT,

so the traspose of (A - aI)x is xT(A - aI)T, which is a different

matrix from (A - aI)Tx.

Here is a proof of the assertion that the questioner wanted

to prove:

Proof: Let A be a matrix with an eigenvalue a. Then the

determinant of the matrix A - aI is zero, i.e. det(A - aI) = 0.

We know that the determinant is transpose invariant, i.e.

det(C) = det(C^t) (here C^t means the transpose of C).

Thus, det(A - aI) = det((A - aI)^t). On the other hand,

(A - aI)^t = A^t - aI^t = A^t - aI (as two commentators remembered).

Thus, det(A^t - aI) = det(A - aI) = 0. Hence, a is also an

eigenvalue of A^t, the transpose of the matrix A. QED

Puggy is absolutely correct. The property of transpose on addition or subtraction is distrubitve!

Why that someone would impose the property of transpose on a matrix multiplication beats me unless they were half asleep writing this.

• Puggy
Lv 7

It looks correct to me as well. I noticed you used the property that

(A + B)^T = A^T + B^T

And this is absolutely correct. What confuses me is how someone claimed you used the property

(AB)^T = (B^T)(A^T)

which you didn't, because you're taking the transpose of two matrices being subtracted (which is a special form of adding).