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The normal matrix ${\bf m}$ has eigenvectors $v_i$ with corresponding eigenvalues $\lambda_i$

$$\lambda_iv_i={\bf m}v_i.\tag{1}$$ $u_i$ is the Gram-Schmidt orthogonalized eigenvector that corresponds to $v_i$. How to show that this also implies $$\lambda_iu_i={\bf m}u_i\tag{2} \,\,\,?$$

Incomplete proof :

Gram-Schmidt orthogonalization can be written as $$u_i=v_i+\sum_{k=1}^{i-1}c_kv_k\quad\text{with}\,\, c_k\in \mathbb{R}.\tag{3}$$ After setting eq.(3) into eq.(1) we get $$\lambda_i\left(u_i-\sum_{k=1}^{i-1}c_kv_k\right)={\bf m}\left(u_i-\sum_{k=1}^{i-1}c_kv_k\right)\tag{4}$$ and it follows $$\lambda_i u_i-{\bf m}u_i=\lambda_i \sum_{k=1}^{i-1}c_kv_k-{\bf m}\sum_{k=1}^{i-1}c_kv_k\tag{5}.$$ It remains to show that $\lambda_i$ is an eigenvalue to the vector $\displaystyle{\sum_{k=1}^{i-1}c_kv_k}$. How can this be done?

granular_bastard
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    In general, this only holds for normal matrices (i.e. matrices for which $\mathbf m ^T\mathbf m = \mathbf m\mathbf m^T)$. – Ben Grossmann Feb 17 '22 at 16:36
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    Note that if $v_i,v_j$ correspond to eigenvalues $\lambda_i\neq\lambda_j$, then $v_i$ and $v_j$ are orthogonal (as is shown in this post for instance). It follows that for certain $k$, the coefficients $c_k$ will be zero, which is enough to conclude that $\sum c_k v_k$ is an eigenvector. – Ben Grossmann Feb 19 '22 at 22:06

1 Answers1

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Since ${\bf m}$ is normal, $v_i$ and $v_j$ are normal if $\lambda_i\ne \lambda_j$.

Consider the start of the Gram-Schmidt process. $$u_1=v_1, u_2=v_2+c_1v_1\quad\text{with}\,\, c_1\in \mathbb{R}.$$ Then, by Gram-Schmidt, $u_1.u_2=0$.

By our opening observation, either $u_1.v_2=0$ and then $c_1=0$ or $\lambda_1= \lambda_2$. In both cases, $\lambda_2u_2={\bf m}u_2 $.

Proceeding in this way through the Gram-Schmidt process we see that each $u_i$ is a linear combination of $v_j$ with the same eigenvector as $v_i$ and so each each $u_i$ has eigenvalue $\lambda_i$.

The general step of Gram-Schmidt

$$u_i=v_i-\sum_1^{i-1}\frac{u_k.v_i}{u_k.u_k}u_k$$ where $u_1, ..., u_{i-1}$ are the previously generated orthogonal eigenvectors. The $u_k$ which correspond to eigenvalues other than $\lambda_i$ satisfy $u_k.v_i=0$ and so the only $u_k$ with non-zero coefficients are those with eigenvalue $\lambda_i$. Therefore $${\bf m}.u_i={\bf m}.v_i-\sum_1^{i-1}\frac{u_k.v_i}{u_k.u_k}{\bf m}.u_k=\lambda_iv_i-\sum_1^{i-1}\frac{u_k.v_i}{u_k.u_k}\lambda_iu_k =\lambda_iu_i $$

  • I've added in the general step –  Feb 23 '22 at 16:20
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    @granularbastard "More equations" does not mean "more mathematical". Most proofs are easier to understand when they use words more and symbols less. – Ethan Bolker Feb 23 '22 at 16:23
  • Is this property standard content of linear algebra textbooks or is the question of the OP more of special interest? – granular_bastard Feb 23 '22 at 17:55
  • I've just checked my (old) version of Herstein and it is not mentioned. I think the reason is that what really matters is obtaining the orthonormal basis - and not the connection between it and the original basis. –  Feb 23 '22 at 18:07
  • Is there also a simple geometrical proof in the sense "proof without words"? – granular_bastard Feb 23 '22 at 19:17
  • Possibly not a 'proof'. However a good 3-D illustration of what is happening is to picture a transformation which leaves the z-axis invariant (i.e $\lambda=1$) but doubles vectors in the $x-y$ plane (i.e $\lambda=2$). Then Gram-Schmidt operates as described above, not affecting the direction of an eigenvector in the $z-$direction but orthogonalising eigenvectors in the $x-y$ plane. –  Feb 23 '22 at 19:25