Let \({y_k} = {k^2}\)and \({z_k} = 2k\left| k \right|\). Are the signals \(\left\{ {{y_k}} \right\}\) and \(\left\{ {{z_k}} \right\}\) linearly independent? Evaluate the associated Casorati matrix \(C\left( k \right)\) for \(k = 0\), \(k = - 1\), and \(k = - 2\), and discuss your results.

It is given that \({y_k} = {k^2}\) and \({z_k} = 2k\left| k \right|\).

\(C\left( k \right)\)can be evaluated as shown below:

\(\begin{aligned} C\left( k \right) &= \left( {\begin{array}{*{20}{c}}{{y_k}}&{{z_k}}\\{{y_{k + 1}}}&{{z_{k + 1}}}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{{k^2}}&{2k\left| k \right|}\\{{{\left( {k + 1} \right)}^2}}&{2\left( {k + 1} \right)\left| {k + 1} \right|}\end{array}} \right)\end{aligned}\)

\(\begin{aligned} C\left( 0 \right) &= \left( {\begin{array}{*{20}{c}}{{{\left( 0 \right)}^2}}&{2\left( 0 \right)\left| 0 \right|}\\{{{\left( {0 + 1} \right)}^2}}&{2\left( {0 + 1} \right)\left| {0 + 1} \right|}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}0&0\\1&1\end{array}} \right)\end{aligned}\)

\(\begin{aligned} C\left( 0 \right) &= \left( {\begin{array}{*{20}{c}}{{{\left( { - 1} \right)}^2}}&{2\left( { - 1} \right)\left| { - 1} \right|}\\{{{\left( { - 1 + 1} \right)}^2}}&{2\left( { - 1 + 1} \right)\left| { - 1 + 1} \right|}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}1&2\\0&0\end{array}} \right)\end{aligned}\)

\(\begin{aligned} C\left( { - 2} \right) &= \left( {\begin{array}{*{20}{c}}{{{\left( { - 2} \right)}^2}}&{2\left( { - 2} \right)\left| { - 2} \right|}\\{{{\left( { - 2 + 1} \right)}^2}}&{2\left( { - 2 + 1} \right)\left| { - 2 + 1} \right|}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}4&{ - 8}\\1&{ - 2}\end{array}} \right)\end{aligned}\)

The determinant of the Casorati matrix \(C\left( k \right)\)is evaluated below:

\(\begin{aligned} C\left( k \right) &= \left( {\begin{array}{*{20}{c}}{{k^2}}&{2k\left| k \right|}\\{{{\left( {k + 1} \right)}^2}}&{2\left( {k + 1} \right)\left| {k + 1} \right|}\end{array}} \right)\\ &= {k^2}\left( {2\left( {k + 1} \right)\left| {k + 1} \right|} \right) - \left( {2k\left| k \right|} \right){\left( {k + 1} \right)^2}\\ &= 2k\left( {k + 1} \right)\left( {k\left| {k + 1} \right| - \left( {k + 1} \right)\left| k \right|} \right)\end{aligned}\)

There are three factors in the determinant expression.

The determinant is 0 if all or either of the three factors is zero.

So, if either \(k = 0\) or \(k = - 1\), the determinant is 0.

If \(k > 0\) or \(k > - 1\), the third factor simplifies as

\(\begin{array}{c}k\left| {k + 1} \right| - \left( {k + 1} \right)\left| k \right| = k\left( {k + 1} \right) - \left( {k + 1} \right)k\\ = 0.\end{array}\)

If \(k + 1 < 0\)or \(k < - 1\), the third factor simplifies as

\(\begin{array}{c}k\left| {k + 1} \right| - \left( {k + 1} \right)\left| k \right| = - k\left( {k + 1} \right) + \left( {k + 1} \right)k\\ = 0.\end{array}\)

For all the natural values \(k\), the value of the determinant of \(C\left( k \right)\)is 0. Thus, the Casorati matrix cannot be inverted. It implies that no information can be extracted about signals \({y_k}\) and \({z_k}\), whether these are linearly independent or not.