Let $P :R^3 \rightarrow R^3$ be an orthogonal projection on the $x+2y+z=0$ plane.
Without calculating the projection matrix, determine its 3 eigenvalues and 3 linearly independent eigenvectors.
[Math] Eigenvalues and eigenvectors of orthogonal projection matrix
linear algebra
Related Solutions
To calculate the projection matrix you need to calculate the projection of the canonical basis $\{ e_1=(1,0,0), e_2=(0,1,0), e_3=(0,0,1)\}$ and we know that the plane $P=\{ax+by+cz=0 \}$ is equal to $Vect\{ v_1=(-b,a,0), v_2=(-c,0,a) \}$ in this case we need an orthonormal basis so we construct it by Gram-Schmidt process and we have $$ P=Vect\{f_1=(\frac{-b}{\sqrt{a^2+b^2}},\frac{a}{\sqrt{a^2+b^2}},0); f_2=(\frac{-a^2c}{\alpha},\frac{-abc}{\alpha},\frac{a^3+ab^2-bc}{\alpha})\} $$ With $\alpha=\sqrt{(a^2c)^2+(abc)^2+(a^3+ab^2-bc)^2}$ a normalisation constant.
So the projection of $X=(x,y,z)$ on $P$ is equal to $<X,f_1>f_1 +<X,f_2>f_2$ so in this case :
$$\left\{ \begin{array}{c} A(e_1)&=&<e_1,f_1>f_1 +<e_1,f_2>f_2&=&\frac{-b}{\sqrt{a^2+b^2}} f_1 -\frac{a^2c}{\alpha} f_2&=&(\frac{b^2}{a^2+b^2}+\frac{(a^2c)^2}{\alpha^2},\frac{-ab}{a^2+b^2}-\frac{a^3 bc^2}{\alpha^2},\frac{-a^2c(a^3+ab^2-bc)}{\alpha^2})\\ A(e_2)&=&<e_2,f_1>f_1 +<e_2,f_2>f_2&=&\frac{a}{\sqrt{a^2+b^2}} f_1-\frac{abc}{\alpha}f_2 &=&(\frac{-ab}{a^2+b^2}+\frac{a^3 bc^2}{\alpha^2},\frac{a^2}{a^2+b^2}+\frac{(abc)^2}{\alpha^2},\frac{-abc(a^3+ab^2-bc)}{\alpha^2})\\ A(e_3)&=&<e_3,f_1>f_1 +<e_3,f_2>f_2&=& \frac{a^3+ab^2-bc}{\alpha} f_2&=&(\frac{-a^2c(a^3+ab^2-bc)}{\alpha^2},\frac{-abc(a^3+ab^2-bc)}{\alpha^2},\frac{(a^3+ab^2-bc)^2}{\alpha^2}) \end{array}\right. $$ and finally the matrix is : $$A=\left( \begin{matrix} \frac{b^2}{a^2+b^2}+\frac{(a^2c)^2}{\alpha^2} & \frac{-ab}{a^2+b^2}+\frac{a^3 bc^2}{\alpha^2}& \frac{-a^2c(a^3+ab^2-bc)}{\alpha^2}\\ \frac{-ab}{a^2+b^2}-\frac{a^3 bc^2}{\alpha^2} &\frac{a^2}{a^2+b^2}+\frac{(abc)^2}{\alpha^2} &\frac{-abc(a^3+ab^2-bc)}{\alpha^2}\\ \frac{-a^2c(a^3+ab^2-bc)}{\alpha^2} &\frac{-abc(a^3+ab^2-bc)}{\alpha^2} &\frac{(a^3+ab^2-bc)^2}{\alpha^2} \end{matrix}\right) $$
For $B$ we know that $B=2A-I_3$ so : $$B=\left( \begin{matrix} \frac{2b^2}{a^2+b^2}+\frac{2(a^2c)^2}{\alpha^2}-1 & \frac{-2ab}{a^2+b^2}+\frac{2a^3 bc^2}{\alpha^2}& \frac{-2a^2c(a^3+ab^2-bc)}{\alpha^2}\\ \frac{-2ab}{a^2+b^2}-\frac{2a^3 bc^2}{\alpha^2} &\frac{2a^2}{a^2+b^2}+\frac{2(abc)^2}{\alpha^2}-1 &\frac{-2abc(a^3+ab^2-bc)}{\alpha^2}\\ \frac{-2a^2c(a^3+ab^2-bc)}{\alpha^2} &\frac{-2abc(a^3+ab^2-bc)}{\alpha^2} &\frac{2(a^3+ab^2-bc)^2}{\alpha^2}-1 \end{matrix}\right) $$
For your first question, the identity matrix does the trick: any two vectors, orthogonal or not, are eigenvectors with eigenvalue 1.
More generally, any combination of two eigenvectors with the same eigenvalue $\lambda$ is itself an eigenvector (with eigenvalue $\lambda$); even if your two original eigenvectors are orthogonal, a linear combinations thereof will not be orthogonal to either one.
For the second question, a complex-valued matrix has real eigenvalues iff the matrix is Hermitian, which is to say that it is equal to the conjugate of its transpose: $A^\dagger = (A^T)^* = A$. So while your $A$ is not Hermitian, the matrix $$ B = \begin{bmatrix} 1 & i \\ -i & 1 \end{bmatrix} $$ is, and has two real eigenvalues (0 & 2).
Best Answer
Let $V$ be the subspace of $\Bbb R^3$ defined by $$ V=\{(x,y,z)\in\Bbb R^3:x+2y+z=0\} $$ Now, note that every vector in $V$ is of the form $$ (-2y-z,y,z)=y(-2,1,0)+z(-1,0,1). $$ Putting $v_2=(-2,1,0)$ and $v_3=(-1,0,1)$ then gives $V=\operatorname{Span}\{v_2,v_3\}$.
The map $P:\Bbb R^3\to \Bbb R^3$ is defined by fixing $V$ and orthogonally projecting the vectors not in $V$ onto $V$.
The vector $v_1=(1,2,1)$ is normal to the given plane and $P$ sends this vector to the origin. Hence $P(v_1)=0\cdot v_1$ so that $v_1$ is an eigenvector of $P$ with eigenvalue $\lambda_1=0$.
Next, since $P$ fixes $V=\operatorname{Span}\{v_2,v_3\}$, we have $P(v_2)=v_2$ and $P(v_3)=v_3$. Hence $v_2$ and $v_3$ are eigenvectors of $P$ with eigenvalue $1$.