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This question already has an answer here:

I have to teach about diagonalization of a matrix to a first year undergrad student and I was wondering what would be a good way to motivate this concept. I would appreciate any suggestions. Thanks!

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marked as duplicate by Hans Lundmark, abel, David K, Claude Leibovici, Alexander Konovalov Mar 22 '15 at 13:44

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If $F_i$ denotes the $i$-th Fibonacci number, then $$\left(\begin{matrix} 1&1\\1&0 \end{matrix}\right)\left(\begin{matrix} F_i\\F_{i-1} \end{matrix}\right)=\left(\begin{matrix} F_{i+1}\\F_{i} \end{matrix}\right)$$ Then diagonalizing $\left(\begin{matrix} 1&1\\1&0 \end{matrix}\right)$ gives a way of deriving the formula for the $n$-th Fibonacci number. (It usually takes me about one class period to get through all the details of the example. Students generally seem to like it.)

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Here is a good example for first year undergrad students: (To see the benefits one only must know elementary properties about matrices)

In the most cases it does take very long time to compute the $n-th$ power of a matrix.

But if a matrix is diagonalizable it is much easier. Here is an example:

If we have $$A=\begin{pmatrix} 1 & 0 \\ 2 & 2 \\ \end{pmatrix}$$

Then we can write

$$A=S^{-1}\underbrace{\begin{pmatrix} 1 & 0 \\ 0 & 2 \\ \end{pmatrix}}_{:=D}S $$

for some invertible matrix $S$.

Then $A^n$ can be easily computed:

$$A^{n}=(S^{-1}DS)^{n}=\underbrace{S^{-1}DS\cdot S^{-1}DS\dots S^{-1}DS}_{n \text{ times }}=S^{-1}D^{n}S=S^{-1}\begin{pmatrix} 1 & 0 \\ 0 & 2^n \\ \end{pmatrix}S$$

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I suppose that you know that a matrix is a representation, in a particular basis, of a linear transformation. Note that the linear transformation is a ''geometric object'' that does not depend by the chose basis, so changing the basis the matrix can change, but the linear transformation is the same.

Now, diagonalizing a matrix is to find the basis in wich such linear transformation is expressed with the simpler possible matrix. This 'basis if formed by the eigenvectors and, in this eigen-basis the matrix is diagonal and has as entries the eigenvalues.

Also, eigenvalues are numbers that characterize the given transformation, and the eigenspaces also (they are represented in different ways for different basis, but geometrically they are allways the same subspaces).

So, in some profound sense, diagonalize a matrix is the way to characterize the transformation in a way that is independent from the basis.

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Understanding the effect of a transformation in $\mathbb{R}^3$.

Moments of inertia (spinning books and all that).

Finding the axes of symmetry of quadric surfaces.

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