Linear Algebra: Question about Inverse of Diagonal Matrices

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  • #1
Master1022
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Homework Statement


Not for homework, but just for understanding.

So we know that if a matrix (M) is orthogonal, then its transpose is its inverse.

Using that knowledge for a diagonalised matrix (with eigenvalues), its column vectors are all mutually orthogonal and thus you would assume that its inverse is its transpose...

However, that is wrong and the inverse is actually just the diagonalised matrix with its non-zero entries reciprocated. I understand the proof, but fail to see why the above logic wouldn't apply (beyond that it wouldn't multiply to the identity matrix)

Can anyone help me see where I have gone wrong in trying to combine the two theorems?
 

Answers and Replies

  • #2
DrClaude
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Simply because the columns of a matrix are orthogonal doesn't mean that the matrix is orthogonal. The columns also have to be unit vectors.
 
  • #3
Master1022
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Simply because the columns of a matrix are orthogonal doesn't mean that the matrix is orthogonal. The columns also have to be unit vectors.
Thank you for your response. So what are the other conditions for orthogonality? Even if you normalised the column vectors, it still wouldn't be equal to the reciprocal.

Nevermind, I realized if you normalised the vectors, it becomes the identity matrix.
 
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  • #4
StoneTemplePython
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Homework Statement


Not for homework, but just for understanding.

So we know that if a matrix (M) is orthogonal, then its transpose is its inverse.

Using that knowledge for a diagonalised matrix (with eigenvalues), its column vectors are all mutually orthogonal [should say orthonormal] and thus you would assume that its inverse is its transpose...

However, that is wrong and the inverse is actually just the diagonalised matrix with its non-zero entries reciprocated. I understand the proof, but fail to see why the above logic wouldn't apply (beyond that it wouldn't multiply to the identity matrix)

Can anyone help me see where I have gone wrong in trying to combine the two theorems?

From what I can tell the issues is you haven't carefully written out the math of what you think you're saying -- you're just writing sentences and in the process confusing the underlying matrix/operator with a similarity transform.

##A = UDU^T##
##A^{-1} = \big(UDU^T\big)^{-1} = \big(U^T\big)^{-1}D^{-1}U^{-1} = UD^{-1} U^T##
 
  • #5
Master1022
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From what I can tell the issues is you haven't carefully written out the math of what you think you're saying -- you're just writing sentences and in the process confusing the underlying matrix/operator with a similarity transform.

##A = UDU^T##
##A^{-1} = \big(UDU^T\big)^{-1} = \big(U^T\big)^{-1}D^{-1}U^{-1} = UD^{-1} U^T##
Thanks for the response. I understand the maths that you have written, but perhaps the area I have gone wrong is this orthogonal / orthonormal confusion. Does an orthogonal matrix need to have normalised column vectors?
 
  • #6
PeroK
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Thanks for the response. I understand the maths that you have written, but perhaps the area I have gone wrong is this orthogonal / orthonormal confusion. Does an orthogonal matrix need to have normalised column vectors?

I thought you had proved this for yourself.
 
  • #7
StoneTemplePython
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Thanks for the response. I understand the maths that you have written, but perhaps the area I have gone wrong is this orthogonal / orthonormal confusion. Does an orthogonal matrix need to have normalised column vectors?

so, quoting this:

So we know that if a matrix (M) is orthogonal, then its transpose is its inverse.

you have ##M## or in my writeup ##U## is orthogonal. Suppose it is square and each column vector is mutually orthogonal.

Then ##U^T U = D## where ##D## is some diagonal matrix... but based on your quote, you want ##D = I##, so what does that tell you about ##\big \Vert \mathbf u_k\big \Vert_2^2## which are the entries in the kth diagonal spot of ##D##?
 

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