Let \( M_3(\mathbb{R}) \) be the vector space of all \( 3 \times 3 \) real matrices over \( \mathbb{R} \) under usual matrix addition and scalar multiplication. Let \( T_3(\mathbb{R}) \) be the set of all \( 3 \times 3 \) real upper triangular matrices. Which one of the following is TRUE?
Show Hint
Compare dim(M3(R)) minus dim(T3(R)) with the dimension of each candidate matrix space; only one option's dimension can match.
The quotient space \( M_3(\mathbb{R})/T_3(\mathbb{R}) \) is isomorphic to the vector space of all \( 3 \times 3 \) real skew symmetric matrices over \( \mathbb{R} \) under usual matrix addition and scalar multiplication.
The quotient space \( M_3(\mathbb{R})/T_3(\mathbb{R}) \) is isomorphic to the vector space of all \( 3 \times 3 \) real symmetric matrices over \( \mathbb{R} \) under usual matrix addition and scalar multiplication.
The quotient space \( M_3(\mathbb{R})/T_3(\mathbb{R}) \) is isomorphic to the vector space of all \( 3 \times 3 \) real lower triangular matrices over \( \mathbb{R} \) under usual matrix addition and scalar multiplication.
The quotient space \( M_3(\mathbb{R})/T_3(\mathbb{R}) \) is isomorphic to the vector space of all \( 3 \times 3 \) real matrices with trace zero over \( \mathbb{R} \) under usual matrix addition and scalar multiplication.
Show Solution
The Correct Option isA
Solution and Explanation
Step 1: Understanding the Concept.
Instead of just subtracting dimensions, let's build an actual basis of the quotient space $M_3(\mathbb{R})/T_3(\mathbb{R})$ by extending a basis of $T_3(\mathbb{R})$ to a basis of $M_3(\mathbb{R})$, and match it with a basis of one of the four candidate spaces.
Step 2: Key Formula or Approach.
Write $E_{ij}$ for the matrix with a $1$ in row $i$, column $j$, and $0$ everywhere else. A basis of $T_3(\mathbb{R})$ (upper triangular, diagonal included) is $\{E_{11}, E_{12}, E_{13}, E_{22}, E_{23}, E_{33}\}$, six matrices in total. Adding the three strictly lower entries $E_{21}, E_{31}, E_{32}$ completes a basis of all nine matrices in $M_3(\mathbb{R})$.
Step 3: Detailed Explanation.
In the quotient space, the cosets of the six basis matrices of $T_3(\mathbb{R})$ all become the zero vector, since each already lies inside $T_3(\mathbb{R})$. So the quotient space is spanned by the remaining three cosets $[E_{21}]$, $[E_{31}]$, $[E_{32}]$, and these three are linearly independent, since no nonzero combination of $E_{21}, E_{31}, E_{32}$ can land inside $T_3(\mathbb{R})$ (which has no entries below the diagonal). This gives $\dim\left(M_3(\mathbb{R})/T_3(\mathbb{R})\right)=3$ directly.
Now define a linear map $\varphi$ from this quotient space to the skew symmetric matrices by sending $[E_{21}] \mapsto E_{21}-E_{12}$, $[E_{31}] \mapsto E_{31}-E_{13}$, and $[E_{32}] \mapsto E_{32}-E_{23}$, extended linearly. Each of $E_{21}-E_{12}$, $E_{31}-E_{13}$, $E_{32}-E_{23}$ is skew symmetric, since its transpose is its own negative, and these three matrices form a basis of the $3$-dimensional space of skew symmetric matrices. So $\varphi$ sends a basis of the quotient to a basis of the skew symmetric matrices, which makes $\varphi$ a linear isomorphism.
Checking the other options the same way: symmetric matrices need a $6$-element basis (3 diagonal plus 3 off diagonal free entries), full lower triangular matrices also need a $6$-element basis, and trace zero matrices need an $8$-element basis (only one condition dropped from nine). None of these can be put in one to one linear correspondence with our $3$-element quotient basis $\{[E_{21}], [E_{31}], [E_{32}]\}$.
Step 4: Final Answer.
The quotient space has a basis of exactly three cosets, and this matches a basis of the skew symmetric matrices through an explicit linear bijection.
\[ \boxed{M_3(\mathbb{R})/T_3(\mathbb{R}) \cong \{\text{skew symmetric } 3\times3 \text{ matrices}\}} \]
