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Suppose to have a symmetric matrix $A$ with rank equal to $a$. Now let $$ A_1 = A + \frac{x}{2} $$

such that $x$ is a real number different from 0.

Is it always true that rank of $A_1 = a+1$? Can you prove that? In which case this will be true?

Roberta
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  • What does it mean to add a real number to a matrix? – quasi Mar 20 '18 at 14:34
  • How do you add a real number to a matrix? And why would the rank change if $x = 0$, since then $A = A_1$? – Hans Engler Mar 20 '18 at 14:34
  • Are you adding a real number to a matrix? How are you doing that? – MaximusFastidiousIrreverence Mar 20 '18 at 14:36
  • Add a real number it means to add the same quantity (x/2) to all the entries in the matrix A – Roberta Mar 20 '18 at 14:37
  • If I understand your question correctly, its premise seems incorrect. For counterexample, let $A$ be the 2x2 identity matrix, whose rank is $a=2$. The rank of $A_1$ in that case (given suitable restrictions such as that $x\neq 0$ and the like) is not $a+1$ but $a$. – thb Mar 20 '18 at 14:37
  • "To add the same quantity ... to all entries?" That's odd. I would have expected it to add the quantity along the main diagonal only. Regrettably, it may be that your question is confused. – thb Mar 20 '18 at 14:39
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    @thb: Actually before the clarification, given the claim, I thought maybe the direct sum was meant (i.e. adding the value as extra diagonal element, with the corresponding non-diagonal elements zero); because then for $x\ne 0$ the claim would have been true. – celtschk Mar 20 '18 at 14:42

1 Answers1

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This is wrong.

Take any invertible symmetric $n\times n$ matrix, hence of maximal rank $n$; since $A+\frac x2$ has the same size, it cannot have a rank larger than that of $A$.

Moreover, if all entries of $A$ were equal to $-\frac x2$, then rk$(A)$ would be $1$ and rk$(A+\frac x2)$ would be $0$. So you see that the rank can decrease as well.

Arnaud Mortier
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