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Let $A$ be a $7\times 7$ matrix satisfying $2A^2-A^4=I$ .If $A$ has two distinct eigenvalues and each eigenvalue has geometric multipicity $3$,then find the number of non-zero entries in the Jordan Canonical Form of $A$.

Since $2A^2-A^4=I$ so it is a annihilating polynomial of $A$. The minimal polynomial of $A$ must divide $2x^2-x^4-1=0\implies (x^2-1)^2=0$.

Since it has two distinct eigenvalues so that will be $-1,1$.

But how to find the size of the Jordan Block corresponding to eigen values $1,-1$.

Its quite clear that the minimal polynomial will be $(x-1)^m(x+1)^n;m,n>1$ since the matrix is not diagonalizable.

Please give some hints on how to solve the problem.

2 Answers 2

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The Jordan form of $A$ will have precisely one $2 \times 2$ Jordan block. The reason is that the number of Jordan blocks is the sum of the geometric multiplicities of all the eigenvalues (in this case, six). Hence, it will have six Jordan blocks and so precisely one of them will have to be a $2 \times 2$ block and all others are $1 \times 1$ blocks.

This implies that the number of non-zero entries in the Jordan form of $A$ is $8$.

  • 0
    is it possible to find the minimal polynomial of the matrix? – 2017-02-27
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    It will be either $(x-1)^2(x+1)$ or $(x-1)(x+1)^2$ depending on whether the $2 \times 2$ block is associated to the eigenvalue $1$ or $-1$. – 2017-02-27
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Each Jordan block contributes one to the geometric multiplicity. So we are told that there are six Jordan blocks. As the matrix is $7\times7$, five of the blocks are $1\times1$, and one of them is $2\times2$.

The number of nonzero entries is then $5+3=8$.

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    Why each block contributes 1 – 2017-02-27
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    It is a basic fact about Jordan blocks: their eigenspace is one-dimensional. If you didn't know that, you definitely need to make the computation to see it for yourself. – 2017-02-27