Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

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!

5 Answers

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  • 1 decade ago
    Favorite Answer

    seems correct to me

  • 1 decade ago

    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?

    *** Addendum ***

    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

  • 1 decade ago

    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
    1 decade ago

    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).

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  • Anonymous
    1 decade ago

    then Ax-ax=0

    => (A-aI)x=0

    what is aI in 2nd step?

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