What is the geometric meaning of a $3 \times 3$ matrix having all three eigenvalues as zero? I have interpretations in mind for $0$, $1$, and $2$ eigenvalues being zero, but what about all of them?
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Rodrigo de Azevedo
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ThanksABundle
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5It is called a nilpotent matrix. One can show that a matrix $A$ is nilpotent iff $A^k = 0$ for some $k$. – copper.hat Dec 03 '14 at 23:25
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2OK, but geometrically, what does this mean? – ThanksABundle Dec 03 '14 at 23:28
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Not exactly sure about a geometric perspective, but from a Jordan form perspective, there are exactly two nilpotent Jordan form matrices and one of these is zero. – copper.hat Dec 03 '14 at 23:42
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Geometrically, having one or more eigenvalues of zero simply means the nullspace is nontrivial, so that the image is a "crushed" a bit, since it is of lower dimension.
Other than the obvious case of having exactly one 0 eigenvalue, there's no way to predict the dimension of the nullspace from the number of zero eigenvalues alone. The dimension of the nullspace is bounded by the multiplicity of zero eigenvalues, however.
For example, as mentioned in the comments, the matrices with all eigenvalues 0 are the nilpotent matrices, and these can have rank anywhere between $0$ and $n-1$
rschwieb
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If 0 is an eigenvalue, then the nullspace is non-trivial and the matrix is not invertible. Therefore all the equivalent statements given by the invertible matrix theorem that apply to only invertible matrices are false.
Geometrically: recall the determinant is geometrically the signed volume of the unit square under linear transformation, and it is also the product of eigenvalues. A zero determinant means the output is not fully $n$-dimensional, for example transforming the unit cube into a 2d parallelogram.
qwr
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