Let α, β be two non-zero real numbers and v₁, v₂ be two non-zero real vectors of size 3 × 1. Suppose that v₁ and v₂ satisfy $v_1^T{v}_2=0,v_1^T{v}_1=1$, and $v_2^T{v}_2=1$. Let A be the 3 × 3 matrix given by:

$\mathrm{A}=\alpha {\mathrm{v}}_1{\mathrm{v}}_1^{\mathrm{T}}+\beta {\mathrm{v}}_2{\mathrm{v}}_2^{\mathrm{T}}$

The eigenvalues of A are _______

Eigenvalues of Matrix A: Detailed Calculation

This problem asks us to determine the eigenvalues of a specific $3\times 3$ matrix A, which is constructed using two non-zero real vectors, $v_1$ and $v_2$, and two non-zero real numbers, $\alpha$ and $\beta$. The vectors $v_1$ and $v_2$ have particular properties: they are orthogonal and are unit vectors. Understanding these vector properties is key to finding the eigenvalues.

Let's first define the given information:

- Two non-zero real numbers: $\alpha ,\beta$.
- Two non-zero real vectors of size $3\times 1$: $v_1,v_2$.
- Vector properties: - $v_1^T{v}_2=0$: This indicates that vectors $v_1$ and $v_2$ are orthogonal. - $v_1^T{v}_1=1$: This means $v_1$ is a unit vector (its magnitude is 1). - $v_2^T{v}_2=1$: This means $v_2$ is also a unit vector (its magnitude is 1).
- The $3\times 3$ matrix A is given by: $A=\alpha v_1{v}_1^T+\beta v_2{v}_2^T$.

To find the eigenvalues of matrix A, we can test if $v_1$ and $v_2$ are eigenvectors. An eigenvector $x$ of a matrix $M$ satisfies the equation $Mx=\lambda x$, where $\lambda$ is the eigenvalue.

**Eigenvector Analysis: Testing $v_1$ **

Let's apply matrix A to vector $v_1$:

$$A{v}_1=(\alpha v_1{v}_1^T+\beta v_2{v}_2^T)v_1$$

Distribute $v_1$ into the expression:

$$A{v}_1=\alpha v_1(v_1^T{v}_1)+\beta v_2(v_2^T{v}_1)$$

Now, substitute the given properties:

- We know that $v_1^T{v}_1=1$.
- We also know that $v_1^T{v}_2=0$. Since $v_2^T{v}_1$ is the transpose of $v_1^T{v}_2$, and $0$ is a scalar, $v_2^T{v}_1=0$.

Substitute these values into the equation:

$$A{v}_1=\alpha v_1(1)+\beta v_2(0)$$

$$A{v}_1=\alpha v_1+0$$

$$A{v}_1=\alpha v_1$$

This shows that $v_1$ is an eigenvector of matrix A, and the corresponding eigenvalue is $\alpha$.

**Eigenvector Analysis: Testing $v_2$ **

Next, let's apply matrix A to vector $v_2$:

$$A{v}_2=(\alpha v_1{v}_1^T+\beta v_2{v}_2^T)v_2$$

Distribute $v_2$ into the expression:

$$A{v}_2=\alpha v_1(v_1^T{v}_2)+\beta v_2(v_2^T{v}_2)$$

Now, substitute the given properties:

- We know that $v_1^T{v}_2=0$.
- We know that $v_2^T{v}_2=1$.

Substitute these values into the equation:

$$A{v}_2=\alpha v_1(0)+\beta v_2(1)$$

$$A{v}_2=0+\beta v_2$$

$$A{v}_2=\beta v_2$$

This shows that $v_2$ is an eigenvector of matrix A, and the corresponding eigenvalue is $\beta$.

Eigenvector Analysis: Identifying the Third Eigenvector

Since $v_1$ and $v_2$ are orthogonal unit vectors in a 3-dimensional space, they span a 2-dimensional subspace. For a $3\times 3$ matrix, we expect three eigenvalues (counting multiplicity). We have found two: $\alpha$ and $\beta$.

Let $v_3$ be a vector that is orthogonal to both $v_1$ and $v_2$. Since $v_1$ and $v_2$ are linearly independent and span a 2D subspace of ${\mathbb{R}}^3$, such a $v_3$ always exists and forms a basis (specifically, an orthonormal basis if $v_3$ is also a unit vector) with $v_1$ and $v_2$.

This means:

- $v_1^T{v}_3=0$
- $v_2^T{v}_3=0$

Now, let's apply matrix A to this third orthogonal vector $v_3$:

$$A{v}_3=(\alpha v_1{v}_1^T+\beta v_2{v}_2^T)v_3$$

Distribute $v_3$ into the expression:

$$A{v}_3=\alpha v_1(v_1^T{v}_3)+\beta v_2(v_2^T{v}_3)$$

Substitute the orthogonality properties:

$$A{v}_3=\alpha v_1(0)+\beta v_2(0)$$

$$A{v}_3=0+0$$

$$A{v}_3=0$$

This can be written as $A{v}_3=0\cdot v_3$. Therefore, $v_3$ is an eigenvector of matrix A, and the corresponding eigenvalue is $0$.

Summary of Eigenvalues

We have found three eigenvalues for the $3\times 3$ matrix A: $\alpha$, $\beta$, and $0$. These are the eigenvalues of matrix A.

The eigenvalues of A are $0,\alpha ,\beta$.