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Find a basis for the space of 2×2 lower triangular matrices.

The main objective of this question is to find the basis space for the lower triangular matrices.

This question uses the concept of basis space. A set of vectors B is referred to as a basis for a vector space V if each element of V can be expressed as a linear combination of finite components of B in a distinct manner.

Expert Answer

In this question, we have to find the basis space for the lower triangular matrices.

Let $ s $ be the set that is of lower triangular matrixes.

\[A \space = \space a \begin{bmatrix}
a & 0\\
b & c
\end{bmatrix} \space \in \space S\]

\[A \space = \space \begin{bmatrix}
1 & 0\\
0 & 0
\end{bmatrix}  \space + \space b  \begin{bmatrix}
0 & 0\\
1 & 0
\end{bmatrix} \space + \space c  \begin{bmatrix}
0 & 0\\
0 & 1
\end{bmatrix} \]

Linear combination of $A$ results in:

\[A \space = \space \begin{bmatrix}
1 & 0\\
0 & 0
\end{bmatrix}  \space , \space   \begin{bmatrix}
0 & 0\\
1 & 0
\end{bmatrix} \space and \space   \begin{bmatrix}
0 & 0\\
0 & 1
\end{bmatrix} \]

And:

\[A \space = \space \begin{bmatrix}
1 & 0\\
0 & 0
\end{bmatrix}  \space , \space   \begin{bmatrix}
0 & 0\\
1 & 0
\end{bmatrix} \space  , \space   \begin{bmatrix}
0 & 0\\
0 & 1
\end{bmatrix} \]

Hence, the basis space for lower triangular matrices is $ B $. The final answer is:

\[B\space = \space \begin{bmatrix}
1 & 0\\
0 & 0
\end{bmatrix}  \space , \space   \begin{bmatrix}
0 & 0\\
1 & 0
\end{bmatrix} \space  , \space   \begin{bmatrix}
0 & 0\\
0 & 1
\end{bmatrix} \]

Numeric Results

The basis space for the lower triangular matrices is:

\[B \space = \space \begin{bmatrix}
1 & 0\\
0 & 0
\end{bmatrix}  \space , \space   \begin{bmatrix}
0 & 0\\
1 & 0
\end{bmatrix} \space  , \space   \begin{bmatrix}
0 & 0\\
0 & 1
\end{bmatrix} \]

Example

What is the basis space for the lower triangular matrices of 2 x 2 and what is the dimension of this space?

In this question, we have to find the basis space for the lower triangular matrices and dimensions for this vector space.

We know that:

\[W \space = \space x \begin{bmatrix}
x & 0\\
y & z
\end{bmatrix} \space \in \space S\]

\[W \space = \space x\begin{bmatrix}
1 & 0\\
0 & 0
\end{bmatrix}  \space + \space y  \begin{bmatrix}
0 & 0\\
1 & 0
\end{bmatrix} \space + \space z \begin{bmatrix}
0 & 0\\
0 & 1
\end{bmatrix} \]

Linear combination of $W$ results in:

\[W \space = \space \begin{bmatrix}
1 & 0\\
0 & 0
\end{bmatrix}  \space , \space   \begin{bmatrix}
0 & 0\\
1 & 0
\end{bmatrix} \space and \space   \begin{bmatrix}
0 & 0\\
0 & 1
\end{bmatrix} \]

And we also know that:

\[X \space = \space \begin{bmatrix}
1 & 0\\
0 & 0
\end{bmatrix}  \space , \space   \begin{bmatrix}
0 & 0\\
1 & 0
\end{bmatrix} \space  , \space   \begin{bmatrix}
0 & 0\\
0 & 1
\end{bmatrix} \]

Hence, the final answer is that the basis space for lower triangular matrices is $ X $. The dimension of this basis space is $ 3 $ because it has basis elements of $ 3 $.

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