91影视

A subset Wof a linear space Vis called a subspace of Vif

(a) W contains the neutral element 0of V.

(b) W is closed under addition (if fand gare in Wthen so is f+g)

(C) W is closed under scalar multiplication (if fis in Wand kis scalar, then kfis in W ).

we can summarize parts b and c by saying that W is closed under linear combinations.

Consider two upper triangular matrix namely,

$$\begin{gathered} A=\left[\begin{array}{ccc}1& 4& -1\\ 0& 2& 5\\ 0& 0& 3\end{array}\right] \\ B=\left[\begin{array}{ccc}4& 1& 3\\ 0& -3& 2\\ 0& 0& -1\end{array}\right] \end{gathered}$$

Find A+B.

$$\begin{gathered} A+B=\left[\begin{array}{ccc}1& 4& -1\\ 0& 2& 5\\ 0& 0& 3\end{array}\right]+\left[\begin{array}{ccc}4& 1& 3\\ 0& -3& 2\\ 0& 0& -1\end{array}\right] \\ A+B=\left[\begin{array}{ccc}5& 5& 2\\ 0& -1& 7\\ 0& 0& 2\end{array}\right] \end{gathered}$$

Thus, the sum of two upper triangular matrices is also an upper triangular matrix, because all other entries will be zero and only the entries in the upper half can be added and remain as non-zero elements.

Similarly, when multiplying an upper triangular matrix with a scalar, only the non-zero elements in the upper half of the diagonal can be multiplied and all other will remain as zeros.

$$\begin{gathered} C=2\left[\begin{array}{ccc}4& 2& 3\\ 0& -3& 1\\ 0& 0& -1\end{array}\right] \\ C=\left[\begin{array}{ccc}8& 4& 6\\ 0& -6& 2\\ 0& 0& -2\end{array}\right] \end{gathered}$$

Hence, the upper triangular matrices are a subspace of ${\mathrm{鈩�}}^{3x3}$.