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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    $\begingroup$ No. Take any any non-orthogonal basis $(v_1,\dots,v_n)$ and define a linear map $A$ on this basis by sending each $v_i$ to $iv_i$. The eigenspaces are the $n$ lines generated by the $v_i$, and these are by construction not ortgogonal. $\endgroup$ – Olivier Bégassat May 8 '12 at 13:38
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    $\begingroup$ Eigenvectors corresponding to different eigenvalues will be orthogonal if the matrix is symmetric. This is part of the real spectral theorem. $\endgroup$ – Dylan Moreland May 8 '12 at 13:38
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    $\begingroup$ The case @Dylan is describing will apply to your study of principal components, since the underlying matrices are symmetric... $\endgroup$ – J. M. ain't a mathematician May 8 '12 at 14:50
  • $\begingroup$ @PL this is clearly not a duplicate of that question - this question is about the general case when $M$ is not necessarily a real symmetric matrix. $\endgroup$ – KReiser Jul 26 '20 at 1:55

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 $v$.

  • $\begingroup$ Thank you very much for your answers. Could anyone state whether they are orthogonal in PCA case? $\endgroup$ – Bober02 May 8 '12 at 13:45
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    $\begingroup$ For PCA, things can always be set up such that the eigenvectors are orthogonal. On the other hand, I would recommend looking at PCA as a singular value decomposition instead of as an eigendecomposition. It's been discussed here on math.SE a number of times; search around. $\endgroup$ – J. M. ain't a mathematician May 8 '12 at 14:48
  • $\begingroup$ What do you mean by "setting up"? Is there some common technique to achive a singular matrix? $\endgroup$ – Bober02 May 9 '12 at 8:12
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    $\begingroup$ My understanding based on en.wikipedia.org/wiki/Principal_component_analysis is that a singular value decomposition (SVD) of $X$ results in three matrices, where the first of them is composed of the eigenvectors of $X^T X$, and is used for PCA. This matrix is always orthogonal. $\endgroup$ – Uri Jun 14 '13 at 13:04
  • $\begingroup$ Beautiful answer. $\endgroup$ – Don Larynx Nov 13 '13 at 15:14

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.


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.


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 $\lambda_1 \ne \lambda_2$.


$AX_1 = \lambda_1 X_1$ 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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