Show that the matrix has less than three lineraly independent eigenvalues       3 10 5 
-2 -3 -4 
3 5 7

Let's consider the matrix \( A \) given by:

\[
A = \begin{pmatrix}
3 & 10 & 5 \\
-2 & -3 & -4 \\
3 & 5 & 7
\end{pmatrix}
\]

To determine how many linearly independent eigenvalues the matrix has, we need to calculate its **characteristic polynomial** and find its roots. The characteristic polynomial \( p(\lambda) \) is defined as:

\[
p(\lambda) = \det(A - \lambda I)
\]

Where \( I \) is the identity matrix and \( \lambda \) is an eigenvalue. Let's compute this step-by-step.

### Step 1: Subtract \( \lambda I \) from matrix \( A \)

We subtract \( \lambda \) from the diagonal elements of \( A \):

\[
A - \lambda I = \begin{pmatrix}
3 - \lambda & 10 & 5 \\
-2 & -3 - \lambda & -4 \\
3 & 5 & 7 - \lambda
\end{pmatrix}
\]

### Step 2: Calculate the determinant \( \det(A - \lambda I) \)

We now compute the determinant of the matrix \( A - \lambda I \).

\[
\det(A - \lambda I) = \det\begin{pmatrix}
3 - \lambda & 10 & 5 \\
-2 & -3 - \lambda & -4 \\
3 & 5 & 7 - \lambda
\end{pmatrix}
\]

Let's compute this determinant. The determinant of a 3x3 matrix is given by:

\[
\det(A - \lambda I) = (a_{11})\det\begin{pmatrix}a_{22} & a_{23}\\a_{32} & a_{33}\end{pmatrix} - (a_{12})\det\begin{pmatrix}a_{21} & a_{23}\\a_{31} & a_{33}\end{pmatrix} + (a_{13})\det\begin{pmatrix}a_{21} & a_{22}\\a_{31} & a_{32}\end{pmatrix}
\]

### Step 3: Solve the characteristic equation

The roots of the characteristic polynomial give us the eigenvalues. We want to see how many distinct eigenvalues we have (i.e., how many linearly independent eigenvalues).

Let me compute this determinant to finish the proof.The characteristic polynomial of the matrix \( A \) is:

\[
p(\lambda) = -\lambda^3 + 7\lambda^2 - 16\lambda + 12
\]

### Step 4: Find the eigenvalues

To determine the eigenvalues, we need to find the roots of this characteristic polynomial:

\[
-\lambda^3 + 7\lambda^2 - 16\lambda + 12 = 0
\]

Let me solve this cubic equation to find the eigenvalues.The eigenvalues of the matrix \( A \) are \( \lambda = 2 \) and \( \lambda = 3 \).

Since we only have **two distinct eigenvalues**, and fewer than three linearly independent eigenvalues, this matrix does indeed have less than three linearly independent eigenvalues.

Would you like a more detailed explanation or have any questions?

Here are five related questions you might find helpful:
1. How do we interpret the geometric multiplicity of an eigenvalue?
2. What is the significance of repeated eigenvalues?
3. How can we determine if a matrix is diagonalizable based on its eigenvalues?
4. How do eigenvalues relate to the rank of a matrix?
5. How does the characteristic polynomial help in determining the matrix's invertibility?

**Tip:** A matrix with fewer linearly independent eigenvalues than its size (here 3x3) may not be diagonalizable, and this can indicate special geometric properties.

Show that the matrix has less than three linearly independent eigenvalues: 3 10 5; -2 -3 -4; 3 5 7.

Explore the solution for proving that the given 3x3 matrix has fewer than three linearly independent eigenvalues. Learn about key linear algebra concepts like characteristic polynomials and solving cubic equations. Suitable for undergraduate students studying linear algebra.

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