Discover Linear Algebra

Eigenvalues.

Determine the eigenvalues.

So we have an eigenvalue \(\lambda_1 = -1\) of algebraic multiplicity \(m_1 = 3\text{,}\) and another eigenvalue \(\lambda_2 = 3\) of algebraic multiplicity \(m_1 = 4\text{.}\)

Compute a basis of the eigenspace.

Only one parameter is required (for \(x_7\)), so \(\dim E_{-1}(A) = 1\) and we can span this space with the vector \(\uvec{p}_1 = (0,0,0,0,0,0,1)\text{.}\)

Notice that \(A \uvec{p}_1 = - \uvec{p}_1\text{,}\) as expected.

Since the algebraic multiplicity of this eigenvalue is \(m_1 = 3\text{,}\) we need to continue with generalized eigenvectors until we hit dimension \(3\text{.}\)

Extend to a basis for the second generalized eigensubspace.

Continue on to the generalized eigensubspace of degree \(2\text{.}\) To do this, we analyze the null space of \(C^2 = \bbrac{(-1)I - A}^2\text{.}\)

We need two parameters (for \(x_6\) and \(x_7\)), so \(\dim E_{-1}^2(A) = 2\text{,}\) and can be spanned by the vectors

That second vector is already our basis vector from \(E_{-1}^1(A)\text{,}\) so take \(\uvec{p}_2 = (1,0,0,0,0,1,0)\text{.}\)

Notice that \(\bbrac{(-1)I - A} \uvec{p}_2\) is a multiple of \(\uvec{p}_1\text{,}\) so \(\bbrac{(-1)I - A} \uvec{p}_2\) is an eigenvector, as desired.

We still don't have dimension equal to algebraic multiplicity, so continue.

Extend to a basis for the third generalized eigensubspace.

Now we move on to analyzing the null space of \(C^3\) (where \(C = (-1)I - A\)).

Now we need three parameters, so finally \(\dim E_{-1}^3(A) = m_1 = 3\text{.}\) Assigning parameters to \(x_5,x_6,x_7\) leads to basis vectors

Again, the last two are actually already our basis for \(E_{-1}^2(A)\text{,}\) so set \(\uvec{p}_3 = (1,-1,0,-1,1,0,0)\text{.}\)

Notice that \(C \uvec{p}_3 = -2 \uvec{p}_1 - \uvec{p}_2\text{,}\) so that \(C \uvec{p}_3\) lies in \(E_{-1}^2(A)\text{,}\) as desired.

Compute a basis of the eigenspace.

Three parameters are required, so \(\dim E_3(A) = 3\) and we are already almost up to our target multiplicity of \(m_2 = 4\text{.}\) Assign parameters to \(x_5, x_6, x_7\text{,}\) leading to basis vectors

Notice that \(A \uvec{q}_j = 3 \uvec{q}_3\text{,}\) as expected.

Since the algebraic multiplicity of this eigenvalue is \(m_2 = 4\text{,}\) we need to continue with generalized eigenvectors until we hit dimension \(4\text{.}\)

Extend to a basis for the second generalized eigensubspace.

Continue on to the generalized eigensubspace of degree \(2\text{.}\) This should be our last step for this eigenvalue, as our dimension must increase past the eigenspace dimension (which was \(3\)), but can't increase past the algebraic multiplicity (which is \(m_2 = 4\)). The second generalized eigensubspace \(E_3^2(A)\) is the null space of \(C^2 = \bbrac{3 I - A}^2\text{.}\)

Now we need four parameters, as expected. Assigning parameters to \(x_4,x_5,x_6,x_7\) leads to basis vectors

The last two vectors are already in our basis for \(E_3(A)\text{.}\) The first two vectors can be combined to give our other basis vector for \(E_3(A)\text{,}\) so that indicates that either one would be independent from our current basis for \(E_3(A)\text{.}\) Let's choose the simplest one of the two.

Notice that

so \(\bbrac{3 I - A} \uvec{q}_4\) is an eigenvector, as desired.

Since we are now at \(\dim E_3^2(A) = m_2 = 4\text{,}\) our analysis of this eigenvalue is complete.