Eigenvalues of the Given Matrix

The given 3 × 3 matrix is:

\( M = \left(\begin{array}{lll}\rm a^2 & \rm a b & \rm a c \\ \rm a b & \rm b^2 & \rm b c \\ \rm a c & \rm b c &\rm c^2\end{array}\right) \)

We can observe that this matrix has a special form. Let's define a column vector \( v = \begin{pmatrix} a \\ b \\ c \end{pmatrix} \). The transpose of this vector is a row vector \( v^T = \begin{pmatrix} a & b & c \end{pmatrix} \).

Now, let's compute the outer product of the vector \(v\) with itself, i.e., \(v v^T\):

\( v v^T = \begin{pmatrix} a \\ b \\ c \end{pmatrix} \begin{pmatrix} a & b & c \end{pmatrix} = \left(\begin{array}{lll} a \cdot a & a \cdot b & a \cdot c \\ b \cdot a & b \cdot b & b \cdot c \\ c \cdot a & c \cdot b & c \cdot c \end{array}\right) = \left(\begin{array}{lll}\rm a^2 & \rm a b & \rm a c \\ \rm a b & \rm b^2 & \rm b c \\ \rm a c & \rm b c &\rm c^2\end{array}\right) \)

This matches the given matrix \(M\). So, \(M = v v^T\).

Properties of \(v v^T\) Matrices

A matrix formed by the outer product of a non-zero vector \(v\) with itself (\(v v^T\)) is a rank-1 matrix. The rank of a matrix is the dimension of the vector space spanned by its columns (or rows). For \(M = v v^T\), the columns are multiples of the vector \(v\): \(a \cdot v\), \(b \cdot v\), and \(c \cdot v\).

A 3 × 3 matrix with rank 1 has specific properties regarding its eigenvalues:

• There is only one non-zero eigenvalue (assuming \(v\) is not the zero vector).
• The remaining \(3-1=2\) eigenvalues are zero.

Calculating the Non-Zero Eigenvalue

For a matrix of the form \(M = v v^T\), the single non-zero eigenvalue is equal to the dot product of the vector \(v\) with itself, which is \(v^T v\). This is also the trace of the matrix \(M\).

Let's calculate \(v^T v\):

\( v^T v = \begin{pmatrix} a & b & c \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = a \cdot a + b \cdot b + c \cdot c = a^2 + b^2 + c^2 \)

So, the single non-zero eigenvalue is \(a^2 + b^2 + c^2\).

Summary of Eigenvalues

Based on the rank-1 property of the matrix \(M = v v^T\), we have found that:

• One eigenvalue is \(a^2 + b^2 + c^2\).
• The other two eigenvalues are 0.

Therefore, the eigenvalues of the matrix M are \(a^2 + b^2 + c^2\), 0, and 0.