Question:
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?
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.
Updated On: Jul 21, 2026
- 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.
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question.
We are told \( M_3(\mathbb{R}) \) is the vector space of all \( 3 \times 3 \) real matrices, and \( T_3(\mathbb{R}) \) is the subspace of all \( 3 \times 3 \) real upper triangular matrices (this includes the diagonal entries). We need to find which named vector space the quotient \( M_3(\mathbb{R})/T_3(\mathbb{R}) \) is isomorphic to.
Step 2: Key Formula or Approach.
For a subspace \( W \) of a finite dimensional vector space \( V \), the quotient space \( V/W \) has dimension \( \dim(V/W) = \dim(V) - \dim(W) \). Also, two finite dimensional real vector spaces are isomorphic if and only if they have the same dimension, so once we know the dimension of the quotient we can match it to the option with the same dimension.
Step 3: Detailed Explanation.
The space \( M_3(\mathbb{R}) \) of all \( 3 \times 3 \) real matrices has dimension \( 9 \), since each of the \( 9 \) entries can be chosen freely.
An upper triangular \( 3 \times 3 \) matrix has entries only on and above the diagonal, giving \( 3+2+1=6 \) free entries, so \( \dim T_3(\mathbb{R}) = 6 \).
So the quotient has dimension
\[ \dim \left( M_3(\mathbb{R})/T_3(\mathbb{R}) \right) = 9 - 6 = 3 \]
Now check the dimension of each candidate space. Skew symmetric \( 3\times3 \) matrices satisfy \( A^T=-A \), which forces the diagonal to be zero and leaves only the \( 3 \) entries strictly above the diagonal free (each one fixes its mirror entry with a minus sign), so their dimension is \( 3 \). Symmetric matrices have \( 3 \) free diagonal entries plus \( 3 \) free off diagonal entries, so dimension \( 6 \). Lower triangular matrices, counted the same way as upper triangular, also have dimension \( 6 \). Matrices with trace zero remove only one linear condition from the \( 9 \)-dimensional space, so their dimension is \( 8 \).
Only the skew symmetric matrices have dimension \( 3 \), matching the quotient space. There is even a concrete isomorphism: every coset \( A+T_3(\mathbb{R}) \) has a unique strictly lower triangular representative \( L \) (found by using elements of \( T_3(\mathbb{R}) \) to cancel the diagonal and upper part of \( A \)), and the map \( L \mapsto L-L^T \) carries this \( 3 \)-dimensional space of strictly lower triangular matrices bijectively and linearly onto the skew symmetric matrices.
Step 4: Final Answer.
Since the quotient space has dimension \( 3 \), matching only the skew symmetric matrices, with an explicit linear isomorphism between the two, the quotient space \( M_3(\mathbb{R})/T_3(\mathbb{R}) \) is isomorphic to the vector space of \( 3\times3 \) real skew symmetric matrices.
\[ \boxed{M_3(\mathbb{R})/T_3(\mathbb{R}) \cong \{\text{skew symmetric } 3\times3 \text{ matrices}\}} \]
We are told \( M_3(\mathbb{R}) \) is the vector space of all \( 3 \times 3 \) real matrices, and \( T_3(\mathbb{R}) \) is the subspace of all \( 3 \times 3 \) real upper triangular matrices (this includes the diagonal entries). We need to find which named vector space the quotient \( M_3(\mathbb{R})/T_3(\mathbb{R}) \) is isomorphic to.
Step 2: Key Formula or Approach.
For a subspace \( W \) of a finite dimensional vector space \( V \), the quotient space \( V/W \) has dimension \( \dim(V/W) = \dim(V) - \dim(W) \). Also, two finite dimensional real vector spaces are isomorphic if and only if they have the same dimension, so once we know the dimension of the quotient we can match it to the option with the same dimension.
Step 3: Detailed Explanation.
The space \( M_3(\mathbb{R}) \) of all \( 3 \times 3 \) real matrices has dimension \( 9 \), since each of the \( 9 \) entries can be chosen freely.
An upper triangular \( 3 \times 3 \) matrix has entries only on and above the diagonal, giving \( 3+2+1=6 \) free entries, so \( \dim T_3(\mathbb{R}) = 6 \).
So the quotient has dimension
\[ \dim \left( M_3(\mathbb{R})/T_3(\mathbb{R}) \right) = 9 - 6 = 3 \]
Now check the dimension of each candidate space. Skew symmetric \( 3\times3 \) matrices satisfy \( A^T=-A \), which forces the diagonal to be zero and leaves only the \( 3 \) entries strictly above the diagonal free (each one fixes its mirror entry with a minus sign), so their dimension is \( 3 \). Symmetric matrices have \( 3 \) free diagonal entries plus \( 3 \) free off diagonal entries, so dimension \( 6 \). Lower triangular matrices, counted the same way as upper triangular, also have dimension \( 6 \). Matrices with trace zero remove only one linear condition from the \( 9 \)-dimensional space, so their dimension is \( 8 \).
Only the skew symmetric matrices have dimension \( 3 \), matching the quotient space. There is even a concrete isomorphism: every coset \( A+T_3(\mathbb{R}) \) has a unique strictly lower triangular representative \( L \) (found by using elements of \( T_3(\mathbb{R}) \) to cancel the diagonal and upper part of \( A \)), and the map \( L \mapsto L-L^T \) carries this \( 3 \)-dimensional space of strictly lower triangular matrices bijectively and linearly onto the skew symmetric matrices.
Step 4: Final Answer.
Since the quotient space has dimension \( 3 \), matching only the skew symmetric matrices, with an explicit linear isomorphism between the two, the quotient space \( M_3(\mathbb{R})/T_3(\mathbb{R}) \) is isomorphic to the vector space of \( 3\times3 \) real skew symmetric matrices.
\[ \boxed{M_3(\mathbb{R})/T_3(\mathbb{R}) \cong \{\text{skew symmetric } 3\times3 \text{ matrices}\}} \]
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