Extension of orthonormal bases
Task number: 4356
For the following matrix \( \mathbf A \) find an orthonormal basis of \( \ker (\mathbf A) \) and then extend it to an orthonormal basis of \( \mathbb R^5 \).
\[ \mathbf A = \begin{pmatrix} 2 & 0 & 3 & -6 & 0 \\ -5 & 0 & 3 & 8 & 0 \\ 7 & 4 & 0 & -14 & 3 \\ -1 & 1 & -12 & 10 & 7 \\ 5 & 7 & -3 & -8 & -1 \end{pmatrix} \]Also find an orthonormal basis of \({\mathcal R} (\mathbf A) \) and then extend it to an orthonormal basis of \( \mathbb R^5 \).
How do the found bases relate to each other?
Solution
By Gaussian elimination we calculate that the matrix \( \mathbf A \) has the rank \( 4 \), therefore \(\dim(\ker(\mathbf A))=1\) and backward substitution we get \(\ker(\mathbf A)=\mathcal L (\{(6{,}0,2{,}3,0)^T\})\).
After normalization, we get the orthonormal basis of the kernel \(X_1=\left\{\left(\frac{6}{7},0,\frac{2}{7},\frac{3}{7},0\right)^T\right\}\).
If we extend \( X_1 \) to a basis of \( \mathbb R^5 \) e.g. by the second to fifth vectors of the standard basis and perform Gram-Schmidt orthonormalization, we get the basis extension \(Y_1=\left\{ \mathbf e^2, \left(\frac{-4}{\sqrt{245}},0,\frac{15}{\sqrt{245}},\frac{-2}{\sqrt{245}},0\right)^T, \left(\frac{-1}{\sqrt{5}},0{,}0,\frac{2}{\sqrt{5}},0\right)^T, \mathbf e^5 \right\}\)
To obtain an orthonormal basis of \( \mathcal R (\mathbf A) \) we apply Gram-Schmidt orthonormalization to the rows of the matrix \( \mathbf A \), thus obtaining the basis \(Y_2=\left\{ \left(\frac{2}{7} , 0 , \frac{3}{7} , \frac{-6}{7} , 0 \right)^T, \left(\frac{-3}{7} , 0 , \frac{6}{7} , \frac{2}{7} , 0 \right)^T, \left(0 , \frac{4}{5} , 0 , 0 , \frac{3}{5} \right)^T, \left(0 , \frac{-3}{5} , 0 , 0 , \frac{4}{5}\right)^T \right\}\).
To extend it to an orthonormal basis of \( \mathbb R^5 \) it is enough to take any vector linearly independent of the vectors of \( Y_2 \) e.g. the vector \( \mathbf e^5 \) and To obtain the Gram-Schmidt orthonormalization from it the extension \(X_2=\left\{\left(\frac{6}{7},0,\frac{2}{7},\frac{3}{7},0\right)^T\right\}\)
Since \( \ker (\mathbf A) \) is the orthogonal complement of \( \mathcal R (\mathbf A) \), \( Y_2 \) is also a basis of \( \mathcal R (\mathbf A) \) as well as \( X_2 \) is a basis of \( \ker (\mathbf A) \) (in our case indeed \( X_1 = X_2 \)).
In fact, it would be possible to take the basis \( \mathcal R (\mathbf A) \) as \(Y_3=\left\{ \left(\frac{2}{7} , 0 , \frac{3}{7} , \frac{-6}{7} , 0 \right)^T, \left(\frac{-3}{7} , 0 , \frac{6}{7} , \frac{2}{7} , 0 \right)^T, \mathbf e^2, \mathbf e^5\right\}\).