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  1. For a square nxn matrix to be diagonalizable it needs to have n linearly independent eigenvectors.

  2. If and only if a matrix is normal can the n eigen-vectors be made to form an orthonormal basis.

  3. n linearly independent vectors can always be made to form a orthonormal basis


Does this mean that only normal matrices are diagonalizable?

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    $\begingroup$ $n$ linearly independent eigenvectors can be orthogonalized to form an orthonormal basis, but will they remain eigenvectors? $\endgroup$ – Rahul Dec 21 '17 at 12:31
  • $\begingroup$ @Rahul No, if you have a bunch of eigenvectors and apply gramm-schmidt the resuting vectors will not be eigenvectors. $\endgroup$ – David C. Ullrich Dec 21 '17 at 15:01
  • $\begingroup$ @DavidC.Ullrich: Exactly. (Unless they are in the same eigenspace.) $\endgroup$ – Rahul Dec 21 '17 at 16:58
  • $\begingroup$ With a symmetric/hermitian matrix, you can gram schmidt any collection of n linearly independent eigenvectors and end up with an orthonormal basis of eigenvectors, since different eigenspaces are orthogonal. $\endgroup$ – Batman Dec 22 '17 at 3:47
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All normal matrices are diagonalizable. Not all diagonalizable matrices are normal.

Try to find an example of a diagonalizable but not normal matrix on your own (say, in 3 x 3 matrices).

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    $\begingroup$ $2\times2$ would be easier. $\endgroup$ – Andreas Blass Dec 21 '17 at 12:48
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I'll deal with complex matrices.

A normal matrix is diagonalizable with a unitary matrix (this is a rather deep result). Conversely, if $A$ is diagonalizable with a unitary matrix, it is normal: indeed, if $A=UDU^H$ with $U^H=U^{-1}$ and $D$ diagonal, then \begin{align} AA^H&=UDU^H(UDU^H)^H=UDU^HUD^HU^H=UDD^HU^H \\ A^HA&=(UDU^H)^HUDU^H=UD^HU^HUDU^H=UD^HDU^H \end{align} and $AA^H=A^HA$ follows from $DD^H=D^HD$ (being the matrix $D$ diagonal).

With the help of the spectral theorem, one can prove that $\mathbb{C}^n$ has an orthonormal basis of eigenvectors of a normal matrix $A$. Conversely, if an orthonormal basis of eigenvectors of a matrix $A$ exists, then $A$ is easily seen to be unitarily diagonalizable, hence normal.


Now the main point. It is true that any basis of $\mathbb{C}^n$ can be seen as an orthonormal basis, but generally with respect to a different inner product than the standard one. The facts above are deal with orthonormality with respect to the standard inner product, which is the one where the canonical basis is orthornormal.

This is probably better seen with an abstract approach. Let $V$ be a finite dimensional inner product space $V$. We define the concept of normal operator: a linear map $T\colon V\to V$ is normal if $T^*T=TT^*$, where $T^*\colon V\to V$ is the adjoint operator satisfying $$ \langle T^*v,w\rangle=\langle v,Tw\rangle $$ for every $v,w\in V$.

If $\mathscr{B}$ is an orthonormal basis of $V$, then the matrix of an operator with respect to $\mathscr{B}$ is normal if and only if the operator is normal. (Prove it.)

It's easy to see that if $A$ is a non normal $n\times n$ matrix, then the operator $T\colon\mathbb{C}^n\to\mathbb{C}^n$ defined by $Tv=Av$ is not normal with respect to the standard inner product on $\mathbb{C}^n$, be $A$ diagonalizable or not.

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All symmetric matrices are diagonalizable, since there exists linearly independent eigenvectors of the matrix. Applying Gram Schimdt, it is possible to get the orthonormal basis vectors and thus real symmetric matrices can be made to be unitarily/orthogonally diagonalizable. Another way of looking at it is: since the symmetric matrices are special classes of normal matrices and normal matrices are unitarily diagonalizable, then so is the real symmetric matrix.

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