The following property is similar to *what we have seen for orthogonal maps*. It is one of the two pillars on which the diagonalizability of symmetric matrices is based.

Let #V# be a real inner product space and #L:V\rightarrow V# a symmetric linear map. If #W# is an *#L#-invariant linear subspace* of #V#, then also #W^\perp# is invariant under #L#.

Take an arbitrary vector #\vec{x}\in W^\perp#. We prove that #L(\vec{x})\in W^\perp# by showing that #\dotprod{L(\vec{x})}{\vec{y}}=0# for all #\vec{y}\in W#: \[\begin{array}{rcl}\dotprod{L(\vec{x})}{\vec{y}}&=&\dotprod{\vec{x}}{L(\vec{y})}\\&&\phantom{xx}\color{blue}{\text{symmetry of }L}\\&=&0\\&&\phantom{xx}\color{blue}{\vec{x}\in W^\perp\text{ and }L(\vec{y})\in W\text{ because of invariance of }W}\end{array}\]

Suppose that #V# is a finite-dimensional inner product space and # L:V\rightarrow V# is a symmetric linear map with a real number #\lambda# as its only complex eigenvalue. Then #L# is equal to the scalar multiplication #\lambda\, I_V#.

To see this, we use some results on *eigenspaces*. We write #W# for the kernel of #L-\lambda I_V#. This linear subspace of #V# is *invariant* under #L#. Assume that #W# is a *proper subspace* of #V#, so #W^\perp# is not the trivial subspace. By the theorem, #W^\perp# is invariant under #L#. In particular, #L# has an eigenvector with eigenvalue #\lambda#. This means that the eigenvector not only belongs to #W^\perp# but also to #W#. This contradicts #W\cap W^\perp = \{\vec{0}\}# (one of the *properties of the orthogonal complement*). We conclude that #W=V#, so #L-\lambda I_V# is the zero map, which proves that #L =\lambda\,I_V#.

In the setting of the theorem, the restriction of #L# to #W# as well as the restriction of #L# to #W^\perp# is again symmetric. Because #L# is completely determined by these two restricted maps (bases of #W# and #W^\perp# together form a basis of #V# due to *properties of the orthogonal complement*), we can break up the study of symmetric linear maps to the study of such maps on generalized eigenspaces. The comment *Eigenspaces* shows that the generalized eigenspaces are in fact eigenspaces, so #L# is complex *diagonalizable*. Below we will see that #L# is even real diagonalizable.

The other pillar on which the diagonalizability of symmetric maps rests, is the fact that all the complex eigenvalues of such a map are real.

Let #V# be a real inner product space with #\dim {V}\lt\infty# and let #L:V\rightarrow V# be a symmetric linear map. Then all roots of the characteristic equation of #L# are real.

Suppose that #\mu# is a non-real root of the characteristic equation. Due to the comment about *2D invariant subspaces* for real linear maps, there is a two-dimensional invariant linear subspace #U# such that #\left.L\right|_U#, the restriction of #L# to #U#, is a map #U\rightarrow U# with characteristic polynomial #(x-\mu)\cdot (x-\overline{\mu})#. Choose an orthonormal basis #\alpha# for #U#. Then \[

L_\alpha=\matrix{

a & b\\

b & c}

\] is a symmetric matrix due to theorem *Symmetric maps and matrices.* Its characteristic polynomial is equal to

\[

\left|\,\begin{array}{cc}

a-\lambda & b\\

b & c-\lambda

\end{array}\,\right|\ =\ \lambda^2-(a+c)\lambda+ac-b^2

\] The *discriminant* of this quadratic polynomial is #(a+c)^2-4ac+4b^2=(a-c)^2+4b^2\geq 0#. Therefore, the two roots are real. This contradicts the fact that the characteristic polynomial equals #(x-\mu)\cdot (x-\overline{\mu})#. We conclude that all roots of the characteristic equation of #L :V\rightarrow V# are real.

Let #V# be a finite-dimensional inner product space and # L:V\rightarrow V# a symmetric linear map. The comment on eigenspaces of the previous theorem taught us that if # L:V\rightarrow V# has a single real number #\lambda# as its only complex eigenvalue, it is equal to the scalar multiplication #\lambda\, I_V#. The current theorem tells us that #L# has real eigenvalues only. By using the aforementioned application for each *generalized eigenspace* of a symmetric linear map #L#, we find that #L# is diagonalizable. *Later* we will give a proof of this fact where it turns out that the coordinate transformation conjugating #L# to a diagonal matrix can be chosen to be orthogonal.

The vector #\left[ 1 , 1 \right] # is an eigenvector of the symmetric matrix \[A=\matrix{3 & -1 \\ -1 & 3 \\ }\] with eigenvalue #2#. Thus, the span of the vector #\left[ 1 , 1 \right] # is invariant under \(A\).

The second eigenvalue of #A# is distinct from #2#.

Determine an eigenvector of #A# corresponding to this eigenvalue.

#\rv{a,b}=# #{\left[ 1 , -1 \right] }#

The orthogonal complement of #\linspan{\left[ 1 , 1 \right] }# in #\mathbb{R}^2# is #1#-dimensional. A spanning vector of it is an eigenvector. Such a spanning vector #\rv{a,b}# can be found by solving the equation

\[{\dotprod{\left[ 1 , 1 \right] }{\rv{a,b}} =0}\] This leads to the equation #a+b=0#. A solution is #\rv{a,b} = \left[ 1 , -1 \right] #.