Let P be a unitary matrix with inverse Q. Remember that P implements a rigid rotation, which preserves angles. The eigen vectors of pdq correspond to the eigen vectors of D. In fact we can run the eigen vectors of D through Q to find the eigen vectors of pdq. This is a rigid rotation, hence the eigen vectors of pdq remain orthogonal.
Since P was arbitrary, pdq is any unitarily diagonalizable matrix. This is also known as a normal matrix, as described in the previous section. Therefore every normal matrix has n orthogonal eigen vectors.
Conversely, if M has n orthogonal eigen vectors, they remain orthogonal when fed through a unitary matrix. Use Schur's theorem to map M onto a unitarily similar triangular matrix T. Now the eigen vectors of T are orthogonal.
We want to prove T is diagonal.
Let z be the last coordinate vector in the standard basis for n space. If we're talking about 3 space, z = 0,0,1. Multiply z by T and get a scale multiple of z. Thus z is an eigen vector.
Let y be any other eigen vector, which is necessarily perpendicular to z. Its last component is 0. In 3 space it might look like 5,17,0 (for example). Now yT is a scale multiple of y, and that means the last component of yT is 0. In other words, the dot product of y with the last column of T yields 0.
However, the above must hold for every eigen vector y perpendicular to z. There are n-1 of these vectors, and they span n-1 space. In fact they are orthogonal. Put them all together to make a nonsingular matrix of dimension n-1, and multiply by the last column of T, and get 0. This means the last column of T, aside from the lower right entry, is the zero vector. We have cleared the last column of T, and it's beginning to look diagonal.
Let y be the next basis vector, 0,0,0,…,1,0. Now y is an eigen vector, and the previous n-2 eigen vectors conspire to clear the panultimate column of T, just as we did earlier, with the rightmost column of T. This continues until T is diagonal, and its eigen vectors are the basis vectors of n space. Therefore our original matrix M is diagonalizable.
Put this all together and we have three equivalent definitions of a normal matrix.