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My question relates to the conditions under which the spectral decomposition of a nonnegative definite symmetric matrix can be performed. That is if $A$ is a real $n\times n$ symmetric matrix with eigenvalues $\lambda_{1},...,\lambda_{n}$, $X=(x_{1},...,x_{n})$ where $x_{1},...,x_{n}$ are a set of orthonormal eigenvectors that correspond to these eigenvalues (i.e. $X$ is an orthogonal matrix), and $\Lambda=\text{diag}(\lambda_{1},...,\lambda_{n})$ then

$A=X\Lambda X'$

is the spectral decomposition of $A$. If we then let $A^{1/2}=X\Lambda^{1/2}X'$, where $\Lambda^{1/2}$ is a square root matrix of $\Lambda$ - i.e. $\Lambda^{1/2}\Lambda^{1/2}=\Lambda$, then $A^{1/2}A^{1/2}=A$. Thus $A^{1/2}$ is a square root matrix of $A$.

So if a real nonnegative definite symmetric $n\times n$ matrix $A$ has $n$ eigenvalues then the matrix has a spectral decomposition and thus a square root matrix too. My question is do all symmetric $n\times n$ invertible matrices have $n$ eigenvalues, and thus a square root matrix? Furthermore it is not clear to me whether the eigenvalues have to be distinct or not.

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    $\begingroup$ All real symmetric matrices are diagonalisable, does that help?? $\endgroup$ – Vishesh Jul 26 '13 at 16:37
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    $\begingroup$ The eigenvalues can be repeated. If the geometric multiplicity of the eigenvalues is the same as the algebraic multiplicity then the matrix can be diagonalized. Not all symmetric matrices have distinct eigenvalues, take the identity. $\endgroup$ – Wintermute Jul 26 '13 at 16:46
  • $\begingroup$ Oh yes, my bad, I totally forgot that. Thanks $\endgroup$ – Vishesh Jul 26 '13 at 16:50
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    $\begingroup$ @dandar: Your question suggests you're asking about matrices with real entries, in which case perhaps $\Lambda$ is also required to be real? If that's correct, it's instructive to look at the $1 \times 1$ case. $\endgroup$ – Andrew D. Hwang Jul 26 '13 at 16:52
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    $\begingroup$ Yes the OP should clarify whether the matrix is real, and (if so) whether the square-root is supposed to be real. On the other hand, all complex matrices have complex square-roots; no symmetry is required for that. $\endgroup$ – GEdgar Jul 26 '13 at 16:56
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All symmetric matrices are diagonalizable, therefore they have $n$ eigenvalues (which don't have to be distinct, by the way), all of which are real. The spectral theorem says:

We can decompose any symmetric matrix $A\in S^n$ using symmetric eigendecomposition: $$ A = \sum_{i=1}^n\lambda_iq_iq_i^T = Q\Lambda Q^T, \qquad \Lambda=diag(\lambda_i,\dots,\lambda_n) $$ where the matrix $Q = [q_1,\dots,q_n]$ is orthogonal (with $Q^TQ=I_n$), and contains the eigenvectors of $A$, while the diagonal matrix $\Lambda$ contains the eigenvalues of A.

The matrix "power rule": $$ A^k = Q\Lambda^k Q^{-1} $$ can be used (with $k<0$ being allowed for invertible matrices, which means there should be no $\lambda_i=0$). Note that if there are negative eigenvalues, $\Lambda^{\frac{1}{2}}$ will become complex.

Note that in the complex case, the transpose operations should be replaced with the Hermitian operation (the conjugate transpose).

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  • $\begingroup$ Thank-you for the reply, and yes I now see that all symmetric matrices are diagonalizable. This is because if the matrix is $n\times n$ then we can always construct a set of $n$ orthonormal eigenvectors - i.e. regardless of whether the eigenvalues are repeated or not. Thus we can always compute the spectral decomposition and hence the square root matrix will exist. $\endgroup$ – dandar Jul 26 '13 at 17:12
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What about the matrix $(-1){}{}{}{}{}{}{}{}{}{}{}$?

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  • $\begingroup$ Thank-you for your response. You have found a flaw in my inital summary of the use of the spectral decomposition to find the square root matrix of $A$. I should have stated that $A$ needs to be nonnegative definite. This ensures all the eigenvalues of $A$ are greater than or equal to $0$ which precludes your counter-example. $\endgroup$ – dandar Jul 26 '13 at 17:31
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For your first question: that you can diagonalize real symmetric matrices is the so-called spectral theorem.

For the second: your argument is general; you did not use that the eigenvalues were distinct.

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  • $\begingroup$ Thanks for the reply. Yes you are right the theory for the spectral theorem does not care about whether or not the $n$ eigenvalues of $A$ are distinct or not. $\endgroup$ – dandar Jul 26 '13 at 17:13
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Matrix square root can be defined in many ways. If you just want $X$ such that $X^2 = A$, you approach is good.

However, the principal square root is defined only for the matrices with no strictly negative eigenvalues and zero being at most nonderogatory eigenvalue (which is unimportant here, since symmetric matrices are diagonalizable).

The importance of the principal square root lies in the fact that it is a unique square root with the spectrum in the open right half-plane. If we extended this to the matrices with the real negative eigenvalues, we would either lose uniqueness, or the "open right half-plane" would have to be replaced by something less nice. Of course, there are reasons to ask for this. Read more in Higham's "Functions of Matrices".

Since you ask about symmetric matrices, your eigenvalues are real, so you can only define principal square root if your matrix has nonnegative eigenvalues, which means it is positive semidefinite. That also means that your square (or any other) root will also be positive semidefinite, which can be easily seen from the eigenvalue (spectral) decomposition.

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  • $\begingroup$ Thank-you for your reply. I had not realised the definition of a square root matrix has many forms. $\endgroup$ – dandar Jul 26 '13 at 17:19
  • $\begingroup$ You can compare it to the square root in the set of the nonnegative real numbers. Principal square root is, for example, $\sqrt{4}=2$, but nothing prevents us to also consider $-2$ as a square root of $4$. We even do so, for example, when solving quadratic equations (the $\pm$ sign before $\sqrt{b^2-4ac}$). Unlike the real numbers, which can be considered matrices with a single element, general matrices have much more elements ($n^2$ of them), so you get a much wider variety of candidates for a square root (basically, all combinations of choices for the square root of the eigenvalues). $\endgroup$ – Vedran Šego Jul 26 '13 at 17:41

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