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Diagonalizability and Invertibility

  1. Mar 26, 2009 #1
    Let T be a linear operator on an n-dimensional vector space V. Am I correct in saying that if T is diagonalizable then T is invertible? My reasoning is that if T is diagonalizable then there is an ordered basis of eigenvectors of T such that T's matrix representation is a diagonal matrix. Obviously the rank of the n*n diagonal matrix is n and so the rank of T is n. Now, from a well known theorem we have nullity(T) + rank(T) = dim(V), and since rank(T) = dim(V), we see that nullity(T) = 0, implying that T is both one-to-one and onto, and so T is invertible.
  2. jcsd
  3. Mar 27, 2009 #2

    matt grime

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    The rank of an n by n diagonal matrix is certainly not always n.

    Diagonalizability is about having n linearly independent eigenvectors, and is nothing to do with invertibility. Invertibility is about eigenvalues.
  4. Mar 27, 2009 #3
    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.
    Last edited: Mar 27, 2009
  5. Mar 27, 2009 #4
    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.

    No, this is wrong. If T is invertible, then dim(V)=dim(W), but the converse is false. The zero-map from V to W is invertible only in the trivial case dim(V)=dim(W)=0.
  6. Mar 27, 2009 #5
    If the zero transformation is from V to V, then the matrix representation of that transformation would be an n*n matrix where every entry in the matrix is 0. So wouldn't the rank of the matrix be 0 since there are no linearly independent column vectors? If this is so, then how can the matrix be invertible if its rank is less than n? I thought an n*n matrix is invertible if and only if its rank is n.
  7. Mar 27, 2009 #6

    matt grime

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    n=0 in the case you're troubled by.
  8. Mar 27, 2009 #7


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    You probably forgot that W was supposed to be the image of T. (or maybe that T was supposed to be surjective)
  9. Mar 28, 2009 #8
    Is my reasoning right here:

    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).
  10. Apr 4, 2011 #9


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    do you mean if invertible then no nonzero eigenvalue? Thanks.
  11. Apr 4, 2011 #10


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    No! He means exactly the opposite" "if invertible then no zero eigenvalue".

    I suspect that was what you meant and just mistyped.
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