Orthogonal Vectors and Subspaces
Length of a vector
Let a vector \(x\) be defined such that \(x\in\R^n\) $$ \text{Length of vector }x=||x||=\sqrt{x_1^2+x_2^2+...x_n^2} $$
Orthogonality
Two vectors \(x\) and \(y\) are said to be orthogonal \((\perp)\), if \(x^Ty=y^Tx=0\)
Tip
- \(0\) vector is orthogonal to every vector \(x\).
- If there are \(k\) mutually orthogonal vectors, then the set of that vectors are linearly independent.
Orthonormal Vectors
Let two vectors \(u,v\in\R^n\).
\(u,v\) are said to be orthonormal, if they are both orthogonal and of unit length. Mathematically,
Orthogonal Subspaces
Let \(U, V\) be vector subspaces.
\(U,V\) are vector subspaces if \(u^Tv=v^Tu=0\), where, \(v\in V,u\in U\)
Orthogonal Subspaces and Fundamental Subspaces
\(R(A)\perp N(A)\)
Let \(x\in N(A)\), i.e., \(Ax=0\) which is equal to $$ \begin{bmatrix} \text{row 1}\ \text{row 2}\ \text{row 3}\ \end{bmatrix} \begin{bmatrix} x_1\ x_2\ x_3 \end{bmatrix}= \begin{bmatrix} 0 \ 0\ 0 \end{bmatrix} $$
The above equation implies \(\text{(row 1)}\perp x_1, \text{(row 2)}\perp x_2, ...\) and so on.
Thus for any \(x\), \(\alpha_1r_1+\alpha_2r_2+...\alpha_nr_n\perp x,\) where \(r_i\) is a \(i^{th}\) row of the matrix.
So \(R(A)\perp N(A)\implies C(A^T)\perp N(A)\)
\(C(A)\perp N(A^T)\)
Try it yourself.
Tip
Follow the previous proof with the Column Space and Left Null space.