- #1

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Is there any other proof other than just solving det(A-λI)=0?

- Thread starter brownman
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- #1

- 13

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Is there any other proof other than just solving det(A-λI)=0?

- #2

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I might argue something like the following: By row operations, a rank 1 matrix may be reduced to a matrix with only the first row being nonzero. The eigenvectors of such a matrix may be chosen to be the ordinary Euclidian basis, in which the eigenvalues become zero's and the 11-component of this reduced matrix. As row operations are invertible, the trace is unchanged, and thus this nonzero eigenvalue equals the trace of the original matrix.

Afterthought: But that is probably erroneous, because even though the row operations are indeed invertible, they do not generally preserve the trace. So the last part of my argument fails.

A better argument seems to be the following: For a rank k matrix there exists a basis in which k of its columns are nonzero, the other ones being zero. The transformation between bases may be chosen to be orthogonal, thus preserving the trace.

Afterthought: But that is probably erroneous, because even though the row operations are indeed invertible, they do not generally preserve the trace. So the last part of my argument fails.

A better argument seems to be the following: For a rank k matrix there exists a basis in which k of its columns are nonzero, the other ones being zero. The transformation between bases may be chosen to be orthogonal, thus preserving the trace.

Last edited:

- #3

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So let's see if there are any nonzero eigenvalues.

If ##A## is a rank one matrix, then all of its columns are scalar multiples of each other. Thus we may write ##A = xy^T## where ##x## and ##y## are nonzero ##n \times 1## vectors.

If ##\lambda## is an eigenvalue of ##A##, then there is a nonzero vector ##v## such that ##Av = \lambda v##. This means that ##(xy^T)v = \lambda v##. By associativity, we may rewrite the left hand side as ##x(y^T v) = \lambda v##.

Note that ##y^T v## is a scalar, and of course ##\lambda## is also a scalar. If we assume ##\lambda \neq 0##, then this means that ##v## is a scalar multiple of ##x##: specifically, ##v = x(y^T v)/\lambda##.

Therefore ##x## itself is an eigenvector associated with ##\lambda##, so we have ##x(y^T x) = \lambda x##, or equivalently, ##x(\lambda - y^T x) = 0##. As ##x## is nonzero, this forces ##\lambda = y^T x##.

All that remains is to recognize that ##y^T x = \sum_{n = 1}^{N} x_n y_n## is the trace of ##A = xy^T##.

- #4

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$$A = \begin{bmatrix}

1 & 1 \\

-1 & -1

\end{bmatrix}$$

is a rank one matrix whose only eigenvalue is 0.

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