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The Characteristic polynomial of a matrix $A\in \mathbb{M}_5(\mathbb{R})$ is given by $x^5+\alpha x^4+ \beta x^3$, where $\alpha $ and $\beta$ are non-zero real numbers. What are possible values of the rank of $A$.

Suppose $\lambda_i$'s are the eigenvalues of $A$ for $1\le i\le 5$. $\alpha =\sum \lambda_i\neq0$, thus not every eigenvalue is zero, which implies that rank is not 0. And $\beta=\sum\lambda_i\lambda_j\neq0 \implies $ rank cannot be 1.

And $\prod \lambda_i=0 \implies$ atleast one eigenvalue is zero $\implies$ rank cannot be 5.

Also we know $\sum \lambda_i\lambda_j\lambda_k\lambda_m=0 \implies $ atleast two eigenvalues are 0 $\implies$ rank cannot be 4.

Thus possible values are 2,3.

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    Are you not over thinking this a bit? The characteristic polynomial factors as $x^3(x^2+\alpha x + \beta)$ and so we easily determine it's eigenvalues and multiplicities. Also, where is the question? – 2017-01-24
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    Does that mean rank is 2? – 2017-01-24
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    Look at the Jordan form. The possible ranks vary from 2-4. – 2017-01-24
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    Sir, I don't have much idea about Jordan forms. – 2017-01-24

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$A$ has the eigenvalue $0$ with algebraic multiplicity $3$. It must have at least one eigenvalue associated with $0$, which would mean that $A$ has a nullity (kernel-dimension) of $1$. On the other hand, it could have up to $3$ linearly independent eigenvectors, which corresponds to a nullity of $3$.

Now, apply the rank-nullity theorem.

The possible ranks are $4,3,2$.

An example with rank $4$: $$ \pmatrix{0&1\\&0&1\\ &&0\\ &&&\lambda_1 \\&&&& \lambda_2} $$