The Unapologetic Mathematician

Mathematics for the interested outsider

Diagonal Matrices

Even better than upper-triangular matrices are diagonal matrices. These ones look like

\displaystyle\begin{pmatrix}\lambda_1&&{0}\\&\ddots&\\{0}&&\lambda_d\end{pmatrix}

Each basis vector is an eigenvector, with the eigenvalues listed down the diagonal. It’s straightforward to show that the sum and product of two diagonal matrices are themselves diagonal. Thus, diagonal matrices form a further subalgebra inside the algebra of upper-triangular matrices. This algebra is just a direct sum of copies of \mathbb{F}, with multiplication defined component-by-component.

Diagonal matrices are especially nice because it’s really easy to see how they act on vectors. Given a diagonal matrix D, break a vector v into its components v=\sum v^ie_i. Multiply each component by the corresponding eigenvalue D(v)=\sum(\lambda_iv^i)e_i. And you’re done! Composing a diagonal matrix with another matrix is also easy. To find DT, just multiply each row of T by the corresponding eigenvalue. To find TD, multiply each column of T by the corresponding diagonal.

So, if we can find a basis for our vector space consisting only of eigenvectors for the transformation T, then with respect to that basis the matrix of T is diagonal. This is as good as we can hope for, and a lot of linear algebra comes down to determining when we can do this.

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February 9, 2009 - Posted by | Algebra, Linear Algebra

4 Comments »

  1. [...] with All Eigenvalues Distinct When will a linear transformation have a diagonal matrix? We know that this happens exactly when we can find a basis of our vector space so that each [...]

    Pingback by Transformations with All Eigenvalues Distinct « The Unapologetic Mathematician | February 10, 2009 | Reply

  2. [...] working over an algebraically closed field) then we can pick a basis of eigenvectors and get a diagonal matrix for [...]

    Pingback by Repeated Eigenvalues « The Unapologetic Mathematician | February 11, 2009 | Reply

  3. [...] We’re now ready to characterize those transformations on complex vector spaces which have a diagonal matrix with respect to some basis. First of all, such a transformation must be normal. If we have a [...]

    Pingback by The Complex Spectral Theorem « The Unapologetic Mathematician | August 10, 2009 | Reply

  4. [...] the entry is in the th row and column. The scalings generate the subgroup of diagonal matrices, which is isomorphic to — independent copies of the group of nonzero elements of under [...]

    Pingback by Elementary Matrices « The Unapologetic Mathematician | August 26, 2009 | Reply


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