A vector which has the property that its product with A is the same as its product with a scalar quantity known as its eigenvalue. This follows the form

\begin{align} A\vec{x} = \lambda\vec{x} \end{align}

To find said eigenvector, one must subtract $\lambda\vec{x}$ from both sides after multiplying the right side of the equation by the identity matrix (essentially multiplying by 1, so it's allowed) to give the eigenvalues matrix positions so subtraction makes sense.

Therefore, equation 1 becomes

\begin{align} A\vec{x}-I\lambda\vec{x} = (A-\lambda I)\vec{x} = \vec{0} \end{align}

If we only allow for the values of lambda that makes the statement true, therefore only attempt to find only eigenvectors, we can use the property that the $det(\vec{0})$ to say that the determinant of the previous equation in equation 2 is also 0. This can be represented by the determinant of the nxn matrix

\begin{align} \begin{vmatrix} a-\lambda&b&c&...&n\\d&e-\lambda&f&...&g\\...\\h&i&j&...&n^2-\lambda \end{vmatrix} = 0 \end{align}

Now to avoid the horrible, horrible algebra, let's just say our matrix A is $\begin{bmatrix} 4&0&3&-1\\0&1&1&2\\0&0&-2&0\\0&0&0&4 \end{bmatrix}$

Following the above, general archetype, we can find

\begin{align} det(\begin{vmatrix}4-\lambda&0&3&-1\\0&1-\lambda&1&2\\0&0&-2-\lambda&0\\0&0&0&4-\lambda \end{vmatrix}) = 0 \end{align}
\begin{align} (4-\lambda)(1-\lambda)(-2-\lambda)(4-\lambda) = 0 \end{align}
\begin{align} \lambda = 4,1,-2,4 \end{align}

We say that the eigenvalue 4 has a multiplicity of 2 because it appears twice.

Similar Matrices

Matrix A is similar to B if there exists an invertible matrix P such that

\begin{equation} A = PBP^{-1} \end{equation}
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