I have a very simple question that can be stated without proof. Are all eigenvectors, of any matrix, always orthogonal? I am trying to understand Principal components and it is cruucial for me to see the basis of eigenvectors.
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$\begingroup$In general, for any matrix, the eigenvectors are NOT always orthogonal. But for a special type of matrix, symmetric matrix, the eigenvalues are always real and the corresponding eigenvectors are always orthogonal.
For any matrix M with n rows and m columns, M multiplies with its transpose, either M*M' or M'M, results in a symmetric matrix, so for this symmetric matrix, the eigenvectors are always orthogonal.
In the application of PCA, a dataset of n samples with m features is usually represented in a n* m matrix D. The variance and covariance among those m features can be represented by a m*m matrix D'*D, which is symmetric (numbers on the diagonal represent the variance of each single feature, and the number on row i column j represents the covariance between feature i and j). The PCA is applied on this symmetric matrix, so the eigenvectors are guaranteed to be orthogonal.
$\endgroup$ $\begingroup$Fix two linearly independent vectors $u$ and $v$ in $\mathbb{R}^2$, define $Tu=u$ and $Tv=2v$. Then extend linearly $T$ to a map from $\mathbb{R}^n$ to itself. The eigenvectors of $T$ are $u$ and $v$ (or any multiple). Of course, $u$ need not be perpendicular to $v$.
$\endgroup$ 6 $\begingroup$In the context of PCA: it is usually applied to a positive semi-definite matrix, such as a matrix cross product, $X ' X$, or a covariance or correlation matrix.
In this PSD case, all eigenvalues, $\lambda_i \ge 0$ and if $\lambda_i \ne \lambda_j$, then the corresponding eivenvectors are orthogonal. If $\lambda_i = \lambda_j$ then any two orthogonal vectors serve as eigenvectors for that subspace.
$\endgroup$ $\begingroup$Not necessarily all orthogonal. However two eigenvectors corresponding to different eigenvalues are orthogonal.
e.g Let $X_1$ and $X_2$ be two eigenvectors of a matrix $A$ corresponding to eigenvalues $\lambda_1 $ and $\lambda_2$ where $\lambda_1 \ne \lambda_2$.
Now,
$AX_1 = \lambda_1 X_1$ and $AX_2 = \lambda_2 X_2$.
Taking Transpose of first,
$ \begin{align} &(AX_1)^T = (\lambda_1 X_1)^T\\ \implies & X_1^TA^T = \lambda_1 X_1^T\\ \implies &X_1^TA^TX_2 = \lambda_1 X_1^T X_2\\ \implies & X_1^T\lambda_2 X_2 = \lambda_1 X_1^T X_2\\ \implies &\lambda_2X_1^T X_2 = \lambda_1 X_1^T X_2\\ \implies &(\lambda_2 - \lambda_1)X_1^T X_2 = 0\\ \end{align} $
Since $\lambda_2 \ne \lambda_1$, $X_1^T X_2 $ must be $0$. This establishes orthogonality of eigenvectors corresponding to two different eigenvalues.
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