I'm told the the purpose of diagonalisation is to bring the matrix in a 'nice' form that allows one to quickly compute with it. However in writing the matrix in this nice diagonal form you have to express it w.r.t. a new eigenvector basis. But you'll probably want the answer of your matrix multiplication written w.r.t. to the original basis, so you'll have to do a not-nice matrix multiplication regardless. Example of what I mean:

I want to compute $Ax$, where $A$ and $x$ are given w.r.t. the standard basis ($\epsilon$). However $A$ is quite large and annoying to compute, so I calculate $A_E$ which is a diagonal matrix written w.r.t. the eigenvector basis. But to compute $Ax$ using this matrix, I still have to compute the following: $$ _\epsilon S_E A_E\ _E S_\epsilon$$ Where the $S$ are basis-transition matrices, and those are quite likely to be at least as ugly as our original $A$, so I don't see what we're gaining here. If anything this seems to be a lot more work.

The only thing I can imagine being easier this way is computing something like $A^{10000}x$ or something, because $A^{10000}$ has a really easy form when $A$ is a diagonal matrix, while this would take forever if $A$ is not a diagonal matrix. But is this really the only purpose of diagonalisation; to compute things like $A^{10000}x$?

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    $\begingroup$ one example is that you can compute the determinate straight forwardly. So is the inverse (if it exists). $\endgroup$
    – Chinny84
    Commented May 15, 2015 at 9:34
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    $\begingroup$ And (as you say) it becomes much easier to compute powers, if one can reduce this to diagonal matrices via $A=SDS^{-1}$, $D$ diagonal. Then $A^n=SD^nS^{-1}$. $\endgroup$ Commented May 15, 2015 at 9:34
  • $\begingroup$ Related. $\endgroup$
    – Git Gud
    Commented May 15, 2015 at 9:36
  • $\begingroup$ @GitGud Oke so it seems that the main purpose is indeed just for computing powers of matrices. I guess that must be more useful/important than I assumed it to be then... $\endgroup$ Commented May 15, 2015 at 9:40
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    $\begingroup$ Further to this; almost diagnal Matrices (my term) called the Jordan Normal Form are incredibley useful. See en.wikipedia.org/wiki/Jordan_normal_form $\endgroup$
    – Autolatry
    Commented May 15, 2015 at 9:42

3 Answers 3


I think, in short, the purpose is more to provide a characterization of the matrix you are interested in, in most cases. A "simple" form such as diagonal allows you to instantly determine rank, eigenvalues, invertibility, is it a projection, etc. That is, all properties which are invariant under the similarity transform, are much easier to assess.

A practical example: principal components is an orthogonal diagonalization which give you important information regarding the independent components (eigenvectors) in a system and how important each component is (eigenvalues) - so it allows you to characterize the system in a way which is not possible in the original data. http://en.wikipedia.org/wiki/Principal_component_analysis

I can't think of a case where diagonalization is used purely as a means to "simplify" calculation as it is computationally expensive - it is more of an end goal in itself.

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    $\begingroup$ With PCA and the Singular Value Decomposition you can chuck out the small singular values and obtain a low rank approximation to some data which is incredibly useful for dimensionality reduction, data compression, global optimization and clustering. $\endgroup$
    – Histograms
    Commented May 15, 2015 at 11:41
  • $\begingroup$ Thanks this is the kind of stuff I was hoping to learn about by asking this. $\endgroup$ Commented May 15, 2015 at 15:15
  • $\begingroup$ But finding the diagonal matrix and the invertible matrix itself involves a lot of computation. If we are using computers, then why not use the original matrix itself? And if we are doing it by hand, then finding the inverse of a matrix is nearly as computationally intensive as finding the determinant or higher powers of the original matrix. I am just learning about diagonalization so my comment might sound naive. Pls correct me if I'm wrong! $\endgroup$
    – PGupta
    Commented Jun 10, 2021 at 5:55

I'll add that while you mention computing integer powers of matrices, diagonalization helps in computing fractional powers and exponentiation. If you wanted to compute the matrix $\exp(\mathbf{A})$ you could either go the slow route and use the Taylor series giving: \begin{align}\mathbf{I}+\mathbf{A}+\frac{\mathbf{A}^2}{2!}+\frac{\mathbf{A}^3}{3!}+\dots+\frac{\mathbf{A}^n}{n!}+\dots\end{align}

or alternatively diagonalize $\mathbf{A}$ and it's as easy as doing $e^\lambda$ for each eigenvalue in the diagonal matrix. This significantly reduces the complexity for matrix exponentiation given a required precision.


Here are some situations where you need to compute the diagonal form.

1) First and foremost, diagonalisation is supposed to be applied to endomorphisms, and not matrices, meaning that a basis may not be given.

Example : consider $E$ the vector space of sequences $(u_n)_n$ such that $u_{n+3}=5u_{n+2} + u_{n+1} - u_{n}$. It is well-known that such a sequence is a linear combinaison of exponentials ($\lambda^n$). This comes from the fact that the operator $(u_n)_n \mapsto (u_{n+1})_n$ is a diagonalisable endomorphism on $E$, and $(\lambda^n)_n$ is the eigen-vector for $\lambda$.

2.1) Given a symetric matrix $A$, this can be viewed as a symetric bilinear form. You may want to know if it is a scalar product, and compute an orthognal basis, so you have to compute its diagonal form (but becareful that you need ${}_{\epsilon}S_{E}$ to be orthogonal).

2.2) Given an ellipse's equation $x^2+3y^2+4xy = 1$, you may want to know its axis.

3) Given a differential equation $X'=A.X$, you may want to know if the solutions go to $0$ or $\infty$ and in which directions.


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