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matt grime

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Diagonalizability is about having n linearly independent eigenvectors, and is nothing to do with invertibility. Invertibility is about eigenvalues.

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Hmmm....a diagonal matrix has all non-diagonal entries as 0, and so each row has only one non-zero entry, this entry belonging to a column whose all other entries are 0, so how could one column of the diagonal matrix be written as a linear combination of the others, if you get what I mean? For example, say we have this 3*3 matrix:

A 0 0

0 B 0

0 0 C

No column could ever be written as a linear combination of the others. It seems to me that this would be the case for all diagonal matrices.

EDIT: It seems like what I am saying about diagonal matrices is true only if there is no zero entry on the principal diagonal. Since the diagonal entries of a diagonal matrix are its eigenvalues, then we could say that if no eigenvalue of T is zero then T is invertible. Is this correct?

I also have another question. I once learned that if T is a linear transformation from V to W (both finite dimensional vector spaces), then T is invertible if and only if dim(V) = dim(W). In the case that W = V, T is a linear operator and obviously we have dim(V) = dim(V). So does this imply that ALL linear operators are invertible? Intuitively it doesn't seem like the zero-transformation is invertible.

A 0 0

0 B 0

0 0 C

No column could ever be written as a linear combination of the others. It seems to me that this would be the case for all diagonal matrices.

EDIT: It seems like what I am saying about diagonal matrices is true only if there is no zero entry on the principal diagonal. Since the diagonal entries of a diagonal matrix are its eigenvalues, then we could say that if no eigenvalue of T is zero then T is invertible. Is this correct?

I also have another question. I once learned that if T is a linear transformation from V to W (both finite dimensional vector spaces), then T is invertible if and only if dim(V) = dim(W). In the case that W = V, T is a linear operator and obviously we have dim(V) = dim(V). So does this imply that ALL linear operators are invertible? Intuitively it doesn't seem like the zero-transformation is invertible.

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Yes, an nxn-matrix (no need for it to be diagonal) is invertible if and only if 0 is not an eigenvalue. This is easy to see from the definition of an eigenvalue.EDIT: It seems like what I am saying about diagonal matrices is true only if there is no zero entry on the principal diagonal. Since the diagonal entries of a diagonal matrix are its eigenvalues, then we could say that if no eigenvalue of T is zero then T is invertible. Is this correct?

No, this is wrong.I also have another question. I once learned that if T is a linear transformation from V to W (both finite dimensional vector spaces), then T is invertible if and only if dim(V) = dim(W). In the case that W = V, T is a linear operator and obviously we have dim(V) = dim(V). So does this imply that ALL linear operators are invertible? Intuitively it doesn't seem like the zero-transformation is invertible.

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matt grime

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n=0 in the case you're troubled by.

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Hurkyl

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You probably forgot thatI also have another question. I once learned that if T is a linear transformation from V to W (both finite dimensional vector spaces), then T is invertible if and only if dim(V) = dim(W).

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In the case that dim(V) = 0 then the empty-set spans V and so V is the zero-space (vector space consisting of only the zero vector). If V has only the zero-vector then obviously it has no eigenvectors of T in it, and so there are no eigenvalues. Thus 0 is not an eigenvalue of the zero transformation that acts on the zero-space and so this transformation is invertible.

Am I right in saying that since n = 0, its matrix representation doesn't exist? (I don't see how you can have a 0 by 0 matrix).

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td21

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do you mean if invertible then no nonzero eigenvalue? Thanks.

Diagonalizability is about having n linearly independent eigenvectors, and is nothing to do with invertibility. Invertibility is about eigenvalues.

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HallsofIvy

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No! He means exactly the opposite" "if invertible then nodo you mean if invertible then no nonzero eigenvalue? Thanks.

I suspect that was what you meant and just mistyped.

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