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A direct sum involving a linear map, null and a span.

Dimension of Null space of two linear mapsProof of direct sum with span and basisDirect sum of column space and null space of a square matrixProving that $V=operatornamenullphi oplus au:ainmathbbF$, where $u$ is not in $operatornamenullphi$If $operatornamerangeT' = operatornamespan(varphi)$, then $operatornamenull T = operatornamenull varphi$numerical linear algebra span/nullNull space of linear maps plus span of vectorNilpotent matrix and direct sumIdempotent linear transformation from $V$ to $V$ is the direct sum of $operatornamerange(T)$ and $operatornamenull(T)$Relation of span on a linear map

0

1

$begingroup$

How can I prove that $V = operatornameNullphioplus U$

Suppose that $phi in mathcalL(V,mathbbF)$ and $u in V$ is not in $operatornameNullphi$.

Let $U= operatornameSpan(u)$.

linear-algebra

share|cite|improve this question

edited 14 hours ago

Xander Henderson

14.9k103555

asked Mar 13 at 18:17

noobiskonoobisko

715

$endgroup$

add a comment | 

0

1

$begingroup$

How can I prove that $V = operatornameNullphioplus U$

Suppose that $phi in mathcalL(V,mathbbF)$ and $u in V$ is not in $operatornameNullphi$.

Let $U= operatornameSpan(u)$.

linear-algebra

share|cite|improve this question

edited 14 hours ago

Xander Henderson

14.9k103555

asked Mar 13 at 18:17

noobiskonoobisko

715

$endgroup$

add a comment | 

$begingroup$

How can I prove that $V = operatornameNullphioplus U$

Suppose that $phi in mathcalL(V,mathbbF)$ and $u in V$ is not in $operatornameNullphi$.

Let $U= operatornameSpan(u)$.

linear-algebra

share|cite|improve this question

edited 14 hours ago

Xander Henderson

14.9k103555

asked Mar 13 at 18:17

noobiskonoobisko

715

$endgroup$

share|cite|improve this question

edited 14 hours ago

Xander Henderson

14.9k103555

asked Mar 13 at 18:17

noobiskonoobisko

715

share|cite|improve this question

edited 14 hours ago

Xander Henderson

14.9k103555

asked Mar 13 at 18:17

noobiskonoobisko

715

edited 14 hours ago

Xander Henderson

14.9k103555

edited 14 hours ago

Xander Henderson

14.9k103555

asked Mar 13 at 18:17

noobiskonoobisko

715

asked Mar 13 at 18:17

noobiskonoobisko

715

1 Answer
1

active

oldest

votes

2

$begingroup$

We need

$V = ker phi oplus textspan( u ); tag 0$

I use the notation

$ker phi = textNull ; phi; tag 1$

given that

$u notin ker phi; tag 2$

we have

$0 ne phi(u) in Bbb F; tag 3$

we may thus define

$v = dfrac1phi(u)u; tag 4$

then

$phi(v) = dfrac1phi(u)phi(u) = 1; tag 5$

now for $y in V$ we have

$phi(y - phi(y)v) = phi(y) - phi(y) phi(v) = phi(y) - phi(y) = 0; tag 6$

that is,

$y - phi(y)v in ker phi, tag 7$

and clearly

$phi(y) v in textspan(v) = textspan(u) tag 8$

in light of (4); we have thus shown that every $y in V$ may be written in the form

$y = y - phi(y) v + phi(y)v in ker phi + textspan(u); tag 9$

to complete the establishment of (0), we merely need to show that

$ker phi cap textspan(u) = 0; tag10$

but if

$z in ker phi cap textspan(u), tag11$

then

$z in ker phi, tag12$

that is,

$phi(z) = 0; tag13$

also,

$z in textspan(u) Longrightarrow exists beta in Bbb F, ; z = beta u; tag14$

for such $z$, by (13),

$beta phi(u) = phi(beta u) = phi(z) = 0, tag14$

thus (3) forces

$beta = 0; tag15$

therefore

$z = beta u = 0; tag16$

hence (10) and, finally, (0) are validated.

share|cite|improve this answer

answered Mar 13 at 22:17

Robert LewisRobert Lewis

48.1k23167

$endgroup$

add a comment | 

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2

$begingroup$

We need

$V = ker phi oplus textspan( u ); tag 0$

I use the notation

$ker phi = textNull ; phi; tag 1$

given that

$u notin ker phi; tag 2$

we have

$0 ne phi(u) in Bbb F; tag 3$

we may thus define

$v = dfrac1phi(u)u; tag 4$

then

$phi(v) = dfrac1phi(u)phi(u) = 1; tag 5$

now for $y in V$ we have

$phi(y - phi(y)v) = phi(y) - phi(y) phi(v) = phi(y) - phi(y) = 0; tag 6$

that is,

$y - phi(y)v in ker phi, tag 7$

and clearly

$phi(y) v in textspan(v) = textspan(u) tag 8$

in light of (4); we have thus shown that every $y in V$ may be written in the form

$y = y - phi(y) v + phi(y)v in ker phi + textspan(u); tag 9$

to complete the establishment of (0), we merely need to show that

$ker phi cap textspan(u) = 0; tag10$

but if

$z in ker phi cap textspan(u), tag11$

then

$z in ker phi, tag12$

that is,

$phi(z) = 0; tag13$

also,

$z in textspan(u) Longrightarrow exists beta in Bbb F, ; z = beta u; tag14$

for such $z$, by (13),

$beta phi(u) = phi(beta u) = phi(z) = 0, tag14$

thus (3) forces

$beta = 0; tag15$

therefore

$z = beta u = 0; tag16$

hence (10) and, finally, (0) are validated.

share|cite|improve this answer

answered Mar 13 at 22:17

Robert LewisRobert Lewis

48.1k23167

$endgroup$

add a comment | 

2

$begingroup$

We need

$V = ker phi oplus textspan( u ); tag 0$

I use the notation

$ker phi = textNull ; phi; tag 1$

given that

$u notin ker phi; tag 2$

we have

$0 ne phi(u) in Bbb F; tag 3$

we may thus define

$v = dfrac1phi(u)u; tag 4$

then

$phi(v) = dfrac1phi(u)phi(u) = 1; tag 5$

now for $y in V$ we have

$phi(y - phi(y)v) = phi(y) - phi(y) phi(v) = phi(y) - phi(y) = 0; tag 6$

that is,

$y - phi(y)v in ker phi, tag 7$

and clearly

$phi(y) v in textspan(v) = textspan(u) tag 8$

in light of (4); we have thus shown that every $y in V$ may be written in the form

$y = y - phi(y) v + phi(y)v in ker phi + textspan(u); tag 9$

to complete the establishment of (0), we merely need to show that

$ker phi cap textspan(u) = 0; tag10$

but if

$z in ker phi cap textspan(u), tag11$

then

$z in ker phi, tag12$

that is,

$phi(z) = 0; tag13$

also,

$z in textspan(u) Longrightarrow exists beta in Bbb F, ; z = beta u; tag14$

for such $z$, by (13),

$beta phi(u) = phi(beta u) = phi(z) = 0, tag14$

thus (3) forces

$beta = 0; tag15$

therefore

$z = beta u = 0; tag16$

hence (10) and, finally, (0) are validated.

share|cite|improve this answer

answered Mar 13 at 22:17

Robert LewisRobert Lewis

48.1k23167

$endgroup$

add a comment | 

$begingroup$

We need

$V = ker phi oplus textspan( u ); tag 0$

I use the notation

$ker phi = textNull ; phi; tag 1$

given that

$u notin ker phi; tag 2$

we have

$0 ne phi(u) in Bbb F; tag 3$

we may thus define

$v = dfrac1phi(u)u; tag 4$

then

$phi(v) = dfrac1phi(u)phi(u) = 1; tag 5$

now for $y in V$ we have

$phi(y - phi(y)v) = phi(y) - phi(y) phi(v) = phi(y) - phi(y) = 0; tag 6$

that is,

$y - phi(y)v in ker phi, tag 7$

and clearly

$phi(y) v in textspan(v) = textspan(u) tag 8$

in light of (4); we have thus shown that every $y in V$ may be written in the form

$y = y - phi(y) v + phi(y)v in ker phi + textspan(u); tag 9$

to complete the establishment of (0), we merely need to show that

$ker phi cap textspan(u) = 0; tag10$

but if

$z in ker phi cap textspan(u), tag11$

then

$z in ker phi, tag12$

that is,

$phi(z) = 0; tag13$

also,

$z in textspan(u) Longrightarrow exists beta in Bbb F, ; z = beta u; tag14$

for such $z$, by (13),

$beta phi(u) = phi(beta u) = phi(z) = 0, tag14$

thus (3) forces

$beta = 0; tag15$

therefore

$z = beta u = 0; tag16$

hence (10) and, finally, (0) are validated.

share|cite|improve this answer

answered Mar 13 at 22:17

Robert LewisRobert Lewis

48.1k23167

$endgroup$

share|cite|improve this answer

answered Mar 13 at 22:17

Robert LewisRobert Lewis

48.1k23167

answered Mar 13 at 22:17

Robert LewisRobert Lewis

48.1k23167

answered Mar 13 at 22:17

Robert LewisRobert Lewis

48.1k23167