Show that the rank of $ n\times n$ symmetric tridiagonal matrix is at least $n-1$, and prove that it has $n$ distinct eigenvalues.
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4It would be beneficial to both you and the ones helping you if you showed your own approach to the problem, briefly – Vishesh Oct 10 '13 at 07:47
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4What about the null matrix (that is tridiagonal and symmetric) ? – user37238 Oct 10 '13 at 08:00
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3Do you mean symmetric tridiagonal matrices with nonzero sub-diagonal elements? – Algebraic Pavel Oct 10 '13 at 10:48
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This is for tridiagonal matrices with nonzero off-diagonal elements.
Let $\lambda$ be an eigenvalue of $A\in\mathbb{R}^{n\times n}$ (which is symmetric tridiagonal with nonzero elements $a_{2,1},a_{3,2},\ldots,a_{n,n-1}$ on the subdiagonal). The submatrix constructed by deleting the first row and the last column of $A-\lambda I$ is nonsingular (since it is upper triangular and has nonzero elements on the diagonal) and hence the dimension of the nullspace of $A-\lambda I$ is 1 (because its rank cannot be smaller than $n-1$ and the nullspace must be nontrivial since $\lambda$ is an eigenvalue). It follows then that the geometric multiplicity is 1 and hence the algebraic multiplicity of $\lambda$ is 1 as well. This holds for any eigenvalue of $A$ and hence they are distinct.
The fact that $\mathrm{rank}(A)\geq n-1$ is just a simple consequence of that ($0$ has also multiplicity 1 if $A$ is singular).
Algebraic Pavel
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Where are you using the fact that the matrix is symmetric, or that the superdiagonal contains no zeroes? Or is there a possible generalization here? (In $\mathbb C$ of course.) – kini Oct 18 '16 at 20:27
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@KeshavKini All you need to make this argument to work is an assumption that $A$ is diagonalizable. – Algebraic Pavel Oct 19 '16 at 08:27
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@AlgebraicPavel Sorry for the delayed comment. Why from geometric multiplicity 1 we conclude that we have algebraic multiplicity 1? Did you use the symmetry property or something else? Could you please explain? – darkmoor Oct 07 '21 at 14:34
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@darkmoor The missing detail is that $A$ is diagonalizable because it is symmetric (spectral theorem). Then, the geometric multiplicities must add up to $n$; as all of them are equal to one, there must be $n$ distinct eigenvalues. – jp48 Oct 12 '21 at 08:35