Given is a symmetric bilinear form $\phi:\mathbb R^4\times\mathbb R^4\to \mathbb R$ with $\phi(v,w)=v^{T}Aw$ and $A=\begin{bmatrix}1&1&1&1\\1&2&2&2\\1&2&-2&0\\1&2&0&-3\end{bmatrix}$

How can I compute the orthogonal complement $U$ of $W=(e_1,e_1+e_2+e_3)$ with respect to $\phi$ and second question; Is $U\cap W$ not the empty set ?

$U=\{v\in\mathbb R^4:\phi(e_1,v)=0,\phi(e_1+e_2+e_3,v)=0\}$ but $e_1+e_2+e_3$ is not in $\mathbb R^4$, it is a matrix ?

can I say $\phi(e_1+e_2+e_3,v)=\phi(e_1,v)+\phi(e_2,v)+\phi(e_3,v)$ and then ?


1 Answer 1


The set $W$ can clearly be expressed as the set of vectors $(a+b,b,b,0)^T$.

Expanding orthogonality relationship

$$\begin{bmatrix}x&y& z& t\end{bmatrix}\begin{bmatrix}1&1&1&1\\1&2&2&2\\1&2& -2& 0&\\1&2&0&-3\end{bmatrix}\begin{bmatrix}a+b\\b\\ b\\ 0\end{bmatrix}=0$$

one gets:


that must be verified for all $a,b$, meaning equivalently that $x, y, z, t$ are solutions of the underdetermined system:

$$\begin{cases} x+y+z+t&=&0\\3x+5y+z+3t&=&0\end{cases}$$

That we can for example express in function of the two first variables, giving:

$$\begin{cases} z=y\\t=-x-2y\end{cases}$$

giving the 2 dimensional vector set:

$$\begin{bmatrix}x\\y\\ z\\ t\end{bmatrix}=\begin{bmatrix}x\\y\\y\\ -x-2y\end{bmatrix}=x\begin{bmatrix}1\\0\\0\\-2\end{bmatrix}+y\begin{bmatrix}0\\1\\1\\-2\end{bmatrix}$$

Regarding your second question, a vector of $U \cap W$ could be written, for certain $a,b,x,y$:


from which you will deduce easily that : $a=b=x=y$ (begin by adding the 4 equations): thus we get the null vector; said otherwise: $U \cap W=\{0\}$ (but not "void": beware).

A remark: Matrix $A$ is not positive definite: it has 2 positive eigenvalues and 2 negative eigenvalues.

  • $\begingroup$ I think that you were blocked by the fact you didn't use coordinates. $\endgroup$
    – Jean Marie
    Apr 7, 2016 at 16:48
  • $\begingroup$ Now it's clear thanks a lot $\endgroup$
    – user1161
    Apr 7, 2016 at 17:03

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