Why is $W^perp$ the null space of the basis of $W$?
$begingroup$
Let $W$ be a subspace of $mathbb{R}^n$ and ${w_1, w_2, cdots, w_k}$ form a basis for $W$. Form the matrix $M$ that has these basis vectors as successive columns. According to my book, $W^perp$ is the null space of $M^T$? Why is this?
I would have thought that $W^perp$ is the null space $M$ because I believe that the vectors that are orthogonal to $W$ will be mapped to the origin by $M$. Am I wrong on this and/or drawing a false conclusion?
linear-algebra
asked Dec 18 '18 at 20:50
K. ClaessonK. Claesson
15210
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add a comment |
$begingroup$
Let $W$ be a subspace of $mathbb{R}^n$ and ${w_1, w_2, cdots, w_k}$ form a basis for $W$. Form the matrix $M$ that has these basis vectors as successive columns. According to my book, $W^perp$ is the null space of $M^T$? Why is this?
I would have thought that $W^perp$ is the null space $M$ because I believe that the vectors that are orthogonal to $W$ will be mapped to the origin by $M$. Am I wrong on this and/or drawing a false conclusion?
linear-algebra
asked Dec 18 '18 at 20:50
K. ClaessonK. Claesson
15210
$endgroup$
add a comment |
$begingroup$
Let $W$ be a subspace of $mathbb{R}^n$ and ${w_1, w_2, cdots, w_k}$ form a basis for $W$. Form the matrix $M$ that has these basis vectors as successive columns. According to my book, $W^perp$ is the null space of $M^T$? Why is this?
I would have thought that $W^perp$ is the null space $M$ because I believe that the vectors that are orthogonal to $W$ will be mapped to the origin by $M$. Am I wrong on this and/or drawing a false conclusion?
linear-algebra
asked Dec 18 '18 at 20:50
K. ClaessonK. Claesson
15210
$endgroup$
Let $W$ be a subspace of $mathbb{R}^n$ and ${w_1, w_2, cdots, w_k}$ form a basis for $W$. Form the matrix $M$ that has these basis vectors as successive columns. According to my book, $W^perp$ is the null space of $M^T$? Why is this?
I would have thought that $W^perp$ is the null space $M$ because I believe that the vectors that are orthogonal to $W$ will be mapped to the origin by $M$. Am I wrong on this and/or drawing a false conclusion?
linear-algebra
linear-algebra
asked Dec 18 '18 at 20:50
K. ClaessonK. Claesson
15210
asked Dec 18 '18 at 20:50
K. ClaessonK. Claesson
15210
asked Dec 18 '18 at 20:50
K. ClaessonK. Claesson
15210
asked Dec 18 '18 at 20:50
K. ClaessonK. Claesson
15210
asked Dec 18 '18 at 20:50
K. ClaessonK. Claesson
15210
15210
add a comment |
add a comment |
1 Answer
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Let $x in W^perp$. Then $(M^Tx)_{i} = w_i cdot x = 0 $, since $x$ is orthogonal to $w_i$. This is seen from the definition of transpose and matrix multiplication. Thus $W^{perp} subseteq N(M^T)$. It isn't necessarily the null space for $M$ since matrix multiplication will lead to the dot product of $x$ and the $i^{mathrm{th}}$ row of $M$, which is comprised of the $i^{mathrm{th}}$ entries of each basis vector, which doesn't really tell us anything about the value.
Showing reverse inclusion: If $y in N(M^T)$, then $w_i cdot y = 0$ for all $i$. Thus, $N(M^T) subseteq W^perp$, and so we have $N(M^T) =W^perp$.
answered Dec 18 '18 at 21:09
Anthony TerAnthony Ter
35116
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$begingroup$
Let $x in W^perp$. Then $(M^Tx)_{i} = w_i cdot x = 0 $, since $x$ is orthogonal to $w_i$. This is seen from the definition of transpose and matrix multiplication. Thus $W^{perp} subseteq N(M^T)$. It isn't necessarily the null space for $M$ since matrix multiplication will lead to the dot product of $x$ and the $i^{mathrm{th}}$ row of $M$, which is comprised of the $i^{mathrm{th}}$ entries of each basis vector, which doesn't really tell us anything about the value.
Showing reverse inclusion: If $y in N(M^T)$, then $w_i cdot y = 0$ for all $i$. Thus, $N(M^T) subseteq W^perp$, and so we have $N(M^T) =W^perp$.
answered Dec 18 '18 at 21:09
Anthony TerAnthony Ter
35116
$endgroup$
add a comment |
$begingroup$
Let $x in W^perp$. Then $(M^Tx)_{i} = w_i cdot x = 0 $, since $x$ is orthogonal to $w_i$. This is seen from the definition of transpose and matrix multiplication. Thus $W^{perp} subseteq N(M^T)$. It isn't necessarily the null space for $M$ since matrix multiplication will lead to the dot product of $x$ and the $i^{mathrm{th}}$ row of $M$, which is comprised of the $i^{mathrm{th}}$ entries of each basis vector, which doesn't really tell us anything about the value.
Showing reverse inclusion: If $y in N(M^T)$, then $w_i cdot y = 0$ for all $i$. Thus, $N(M^T) subseteq W^perp$, and so we have $N(M^T) =W^perp$.
answered Dec 18 '18 at 21:09
Anthony TerAnthony Ter
35116
$endgroup$
add a comment |
$begingroup$
Let $x in W^perp$. Then $(M^Tx)_{i} = w_i cdot x = 0 $, since $x$ is orthogonal to $w_i$. This is seen from the definition of transpose and matrix multiplication. Thus $W^{perp} subseteq N(M^T)$. It isn't necessarily the null space for $M$ since matrix multiplication will lead to the dot product of $x$ and the $i^{mathrm{th}}$ row of $M$, which is comprised of the $i^{mathrm{th}}$ entries of each basis vector, which doesn't really tell us anything about the value.
Showing reverse inclusion: If $y in N(M^T)$, then $w_i cdot y = 0$ for all $i$. Thus, $N(M^T) subseteq W^perp$, and so we have $N(M^T) =W^perp$.
answered Dec 18 '18 at 21:09
Anthony TerAnthony Ter
35116
$endgroup$
Let $x in W^perp$. Then $(M^Tx)_{i} = w_i cdot x = 0 $, since $x$ is orthogonal to $w_i$. This is seen from the definition of transpose and matrix multiplication. Thus $W^{perp} subseteq N(M^T)$. It isn't necessarily the null space for $M$ since matrix multiplication will lead to the dot product of $x$ and the $i^{mathrm{th}}$ row of $M$, which is comprised of the $i^{mathrm{th}}$ entries of each basis vector, which doesn't really tell us anything about the value.
Showing reverse inclusion: If $y in N(M^T)$, then $w_i cdot y = 0$ for all $i$. Thus, $N(M^T) subseteq W^perp$, and so we have $N(M^T) =W^perp$.
answered Dec 18 '18 at 21:09
Anthony TerAnthony Ter
35116
answered Dec 18 '18 at 21:09
Anthony TerAnthony Ter
35116
answered Dec 18 '18 at 21:09
Anthony TerAnthony Ter
35116
answered Dec 18 '18 at 21:09
Anthony TerAnthony Ter
35116
35116
add a comment |
add a comment |
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